CBSE Class 11 Maths Important Questions - Chapter 9 Straight Lines

Very Short Answer Questions. (1 Mark)

1. Find the slope of the lines passing through the point $\left(3,-2\right)$ and $\left(-1,4\right)$

Ans: Slope of the line = $m$

The given points are $\left(3,-2\right)$ and $\left(-1,4\right)$.

Let $\left(x_1,y_1\right)$ be $\left(3,-2\right)$ and $\left(x_2,y_2\right)$ be $\left(-1,4\right)$.

$m={\displaystyle \frac{y_2-y_1}{x_2-x_1}}$

$\Rightarrow {\displaystyle \frac{4-\left(-2\right)}{-1-3}}$

$\Rightarrow {\displaystyle \frac{6}{-4}}=-{\displaystyle \frac{3}{2}}$ 

Therefore, the slope of the line passing through the point $\left(3,-2\right)$ and $\left(-1,4\right)$ is $-{\displaystyle \frac{3}{2}}$.

2. Three points $P\left(h,k\right),Q\left(x_1,y_1\right)$ and $R\left(x_2,y_2\right)$ lie on a line. Show that 

$\left(h-x_1\right)\left(y_2-y_1\right)=\left(k-y_1\right)\left(x_2-x_1\right)$

Ans: The given points are $P\left(h,k\right),Q\left(x_1,y_1\right)$and $R\left(x_2,y_2\right)$.

Since, the points $P,Q$ and $R$ lie on a line. Therefore, they are collinear.

Hence,

Slope of $PQ$= Slope of $QR$

$\Rightarrow {\displaystyle \frac{y_1-k}{x_1-h}}={\displaystyle \frac{y_2-y_1}{x_2-x_1}}$                           $\left[\because \text{Slope }={\displaystyle \frac{y_2-y_1}{x_2-x_1}}\right]$

$\Rightarrow {\displaystyle \frac{\left(k-y_1\right)}{\left(h-x_1\right)}}={\displaystyle \frac{y_2-y_1}{x_2-x_1}}$                        (On cross multiplication)

$\left(h-x_1\right)\left(y_2-y_1\right)=\left(k-y_1\right)\left(x_2-x_1\right)$

3. Write the equation of the line through the points $\left(1,-1\right)$ and $$\left(3,5\right)$$

Ans: We know that equation of line through two points $(x_1,y_1)\mathrm{\&}(x_2,y_2)$ is

$y-y_1={\displaystyle \frac{y_2-y_1}{x_2-x_1}}(x-x_1)$

Since the equation of the line through the points $(1,-1)$ and $(3,5)$.

Here,

$x_1=1,y_1=-1$

$x_2=3\text{ and }{y}_2=5$

Putting values

$(y-(-1))={\displaystyle \frac{5-(-1)}{3-1}}(x-1)$

$y+1={\displaystyle \frac{5+1}{2}}(x-1)$

$y+1={\displaystyle \frac{6}{2}}(x-1)$

$y+1=3(x-1)$

$y+1=3x-3$

$y-3x+1+3=0$

$y-3x+4=0$

Hence the required equation is $\mathrm{y}-3\mathrm{x}+4=0$

4. Find the measure of the angle between the lines $x+y+7=0$ and $x-y+1=0$.

Ans: The given equation of lines are $x+y+7=0$ and $x-y+1=0$.

Express the given equation as slope-intercept form $y=mx+c$

where,

(Slope)$m$= coefficient of $x$

$\mathrm{x}+\mathrm{y}+7=0$

${\mathrm{m}}_1={\displaystyle \frac{-1}{1}}$

$\mathrm{x}-\mathrm{y}+1=0$

${\mathrm{m}}_2={\displaystyle \frac{-1}{-1}}=1$

Product of these two slopes is $-1$, therefore, the lines are at right angles.

5. Find the equation of the line that has $y-$intercept $4$ and is perpendicular to the line $y=3x-2$.

Ans: The given equation of the is $y=3x-2$.

Express the given equation as slope-intercept form $y=mx+c$

where,

(Slope)$m$= coefficient of $x$

$m_1=3$

When the lines are perpendicular. Then the product of slope is $-1$.

$\therefore m_1.m_2=-1$

$3.m_2=-1$

$m_2=-{\displaystyle \frac{1}{3}}$

Given,

$y-$intercept of other line is 4.

Therefore, the required equation of the line using the slope-intercept form $y=mx+c$.

$y=-{\displaystyle \frac{1}{3}}x+4$

6. Find the equation of the line, which makes intercepts $-3$ and 2 on the $x$ and $y$-axis respectively.

Ans: Given,

($x-$intercept) a $=-3$

($y-$intercept) b $=2$

Required equation is given by ${\displaystyle \frac{x}{a}}+{\displaystyle \frac{y}{b}}=1$

$a=-3,b=2$

$\therefore {\displaystyle \frac{x}{-3}}+{\displaystyle \frac{y}{2}}=1$

$2x-3y+6=0$

7. Equation of a line is $3x-4y+10=0$ find its slope.

Ans: Given,

Equation of the line is $3x-4y+10=0$.

Slope of the line is given by $m={\displaystyle \frac{-\text{co - efficient of }x}{\text{co - efficient of }y}}$

$\Rightarrow {\displaystyle \frac{-3}{-4}}={\displaystyle \frac{3}{4}}$

8. Find the distance between the parallel lines $3x-4y+7=0$ and $3x-4y+5=0$.

Ans:Given,

Equations of the parallel lines are $3x-4y+7=0$ and $3x-4y+5=0$.

The general equation of the parallel lines is given by $Ax+By+C_1=0\text{ and }Ax+By+C_2=0$.

On comparing the given equation of two parallel lines with general equation, we get

$A=3,B=-4,C_1=7\text{and }{C}_2=5$. 

Distance between two parallel lines is $d={\displaystyle \frac{\left|C_1-C_2\right|}{\sqrt{A^2+B^2}}}$

$\Rightarrow {\displaystyle \frac{|7-5|}{\sqrt{{(3)}^2+{(-4)}^2}}}$

$\Rightarrow {\displaystyle \frac{2}{\sqrt{9+16}}}$

$\Rightarrow {\displaystyle \frac{2}{5}}$

9. Find the equation of a straight line parallel to the $y$-axis and passing through the point $\left(4,-2\right)$.

Ans: Given point is $\left(4,-2\right)$

Equation of line parallel to $y$-axis is $x=a...(i)$

Since, equation (i) passing through $\left(4,-2\right)$

$\therefore a=4$

So,

$x=4$

$x-4=0$

10. If $3x-by+2=0$ and $9x+3y+a=0$ represent the same straight line, find the values of $a$ and $b$.

Ans: Given the equations of the line are $3x-by+2=0$ and $9x+3y+a=0$. 

According to the question,

${\displaystyle \frac{3}{9}}={\displaystyle \frac{-b}{3}}={\displaystyle \frac{2}{a}}$

On comparing the values, we get

${\displaystyle \frac{3}{9}}={\displaystyle \frac{-b}{3}}\text{and }{\displaystyle \frac{3}{9}}={\displaystyle \frac{2}{a}}$

$\therefore b=-1\text{and }a=6$

11. Find the distance between $P\left(x_1{y}_1\right)$ and $Q\left(x_2,y_2\right)$ when $\mathbf{PQ}$ is parallel to the $y$-axis.

Ans. When PQ is parallel to the $y$-axis,

Then $x_1=x_2$

$\therefore PQ=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

$=\sqrt{{\left(x_2-x_2\right)}^2+{\left(y_2-y_1\right)}^2}$

$=\left|y_2-y_1\right|$

12. Find the slope of the line, which makes an angle of ${30}^{\circ }$ with the positive direction of $y-$axis measured anticlockwise.

Ans: Given,

Line which makes an angle of ${30}^{\circ }$ with the positive direction of $y-$axis measured anticlockwise.

Let $\theta$ be the inclination of the line,

$\theta ={120}^{\circ }$

(slope) $m=\tan {120}^{\circ }$

$=\tan \left(90+30\right)$

$=-\sqrt{3}$

13. Determine $x$ so that the inclination of the line containing the points $(x,-3)$ and $(2,5)$ is ${135}^{\circ }$.

Ans: We have been given a line joining the points $(x,-3)$ and $(2,5)$ which makes an angle of ${135}^{\circ }$ with the $x$-axis.

We know that the slope of a line joining the two points $(x_1,y_1)$ and $(x_2,y_2)$ is equal to the tangent of the angle made by the line with $x$-axis in anticlockwise direction given by as follows: slope $=\tan \theta ={\displaystyle \frac{y_2-y_1}{x_2-x_1}}$

So we have $\theta ={135}^{\circ },x_1=2,x_2=x,y_1=5,y_2=-3$

$\Rightarrow \tan {135}^{\circ }={\displaystyle \frac{-3-5}{x-2}}$

Since we know that $\tan {135}^{\circ }=\tan ({90}^{\circ }+{45}^{\circ })=-\cot {45}^{\circ }=-1$ as in second quadrant tangent function is negative.

$\Rightarrow -1={\displaystyle \frac{-3-5}{x-2}}$

$\Rightarrow -1={\displaystyle \frac{-8}{x-2}}$

On cross multiplication, we get as follows:

$\Rightarrow -x+2=-8$

On adding $-2$ to both the sides of the equation, we get as follows:

$\Rightarrow -x-2+2=-8-2$

$\Rightarrow -x=-10$

$\Rightarrow x=10$

Therefore the value of $\mathrm{x}$ is equal to 10 .

14. Find the distance of the point $(4,1)$ from the line $3x-4y-9=0$

Ans: Given,

Point $(4,1)$ and equation of the line $3x-4y-9=0$

Let $\text{d}$ be the required distance,

$d={\displaystyle \frac{|3\cdot (4)-4(1)-9|}{\sqrt{{(3)}^2+{(-4)}^2}}}$

$={\displaystyle \frac{|-1|}{5}}={\displaystyle \frac{1}{5}}$

15. Find the value of $x$ for which the points $(x,-1),(2,1)$ and $(4,5)$ are collinear.

Ans. Given three collinear points.

Let the point be $A(x,-1),B(2,1),C(4,5)$

Since, the points are collinear. Therefore,

Slope of $\text{AB}=$ Slope of $\text{BC}$

${\displaystyle \frac{1+1}{2-x}}={\displaystyle \frac{5-1}{4-2}}$

${\displaystyle \frac{2}{2-x}}={\displaystyle \frac{4}{2}}$

$2-x=1$

$-x=-1$

$x=1$

16. Find the angle between the $x$-axis and the line joining the points $(3,-1)$ and $\left(4,-2\right)$.

Ans: Given,

Line joining the points $(3,-1)$ and $\left(4,-2\right)$.

$m_1=0$ Slope of $x$-axis

$m_2=$ slope of line joining points $(3,-1)$ and $(4,-2)$

$m_2={\displaystyle \frac{y_2-y_1}{x_2-x_1}}$

$={\displaystyle \frac{-2-(-1)}{4-3}}=-1$

Angle between two line is given by $\tan \theta =\left|{\displaystyle \frac{m_2-m_1}{1+m_1{m}_2}}\right|$

$\tan \theta =\left|{\displaystyle \frac{-1+0}{1+0\times (-1)}}\right|$

$\tan \theta =1$

$\theta ={45}^{\circ }$

17. Using slopes, find the value of $x$ for which the points $(x,-1),(2,1)$ and $(4,5)$ are collinear.

Ans: Given,

Three collinear points.

Let the points be A$\left(x,-1\right)$, B$\left(2,1\right)$ and C $\left(4,5\right)$.

Since, the points are collinear. Therefore,

Slope of $\text{AB}=$ Slope of $\text{BC}$

$\Rightarrow {\displaystyle \frac{1-\left(-1\right)}{2-x}}={\displaystyle \frac{5-1}{4-2}}$

$\Rightarrow {\displaystyle \frac{1+1}{2-x}}={\displaystyle \frac{4}{2}}$

$\Rightarrow 2=2\left(2-x\right)$

$\Rightarrow 1=2-x$

$\Rightarrow x=1$

18. Find the value of $K$ so that the line $2x+ky-9=0$ may be parallel to $3x-4y+7=0$

Ans: Given,

The Equations of two parallel lines are $2x+ky-9=0$ and $3x-4y+7=0$.

Slope of the line is given by $m={\displaystyle \frac{-\text{co - efficient of }x}{\text{co - efficient of }y}}$

According to the question,

Slope of 1st line = slope of 2nd line

${\displaystyle \frac{-2}{k}}={\displaystyle \frac{-3}{-4}}$

$\Rightarrow k={\displaystyle \frac{-8}{3}}$

19. Find the value of $K$, given that the distance of the point $(4,1)$ from the line $3x-4y+K=0$ is 4 units.

Ans: Given,

 The distance of the point $(4,1)$ from the line $3x-4y+k=0$ is 4 units.

$\Rightarrow {\displaystyle \frac{|3(4)-4(1)+k|}{\sqrt{{(3)}^2+{(-4)}^2}}}=4$

$\Rightarrow {\displaystyle \frac{\left|12-4+k\right|}{\sqrt{25}}}=4$

$\Rightarrow {\displaystyle \frac{\left|8+k\right|}{5}}=4$

$\Rightarrow \left|8+k\right|=20$

When,

$8+k=20$

  $k=12$

When,

$-\left(8+k\right)=20$

$k=-28$

20. Find the equation of the line through the intersection of $3x-4y+1=0$ and $5x+y-1=0$ which cuts off equal intercepts on the axes.

Ans: Given,

Equation of two intersecting lines $3x-4y+1=0$ and $5x+y-1=0$.

Slope of a line which makes equal intercept on the axes is $-1$ any line through the intersection of given lines is,

$(3x-4y+1)+K(5x-y-1)=0$

$(3+5K)x+y(K-4)+1-K=0$

Slope of the line is given by $m={\displaystyle \frac{-\text{co - efficient of }x}{\text{co - efficient of }y}}$

$\Rightarrow m=-{\displaystyle \frac{(3+5K)}{K-4}}$

$\Rightarrow -1=-{\displaystyle \frac{(3+5K)}{K-4}}$

$\Rightarrow K={\displaystyle \frac{-7}{4}}$

21. Find the distance of the point $(-2,3)$ from the line $12x-5y=2$.

Ans: Given point $(-2,3)$ and equation of the line $12x-5y=2$.

Since, we know

Distance of a point $({\mathrm{x}}_1,{\mathrm{y}}_1)$ from the line $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ is ${\displaystyle \frac{|{\mathrm{ax}}_1+{\mathrm{by}}_1+\mathrm{c}|}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}}$ Distance of the point $(-2,3)$ from the line $12\mathrm{x}-5\mathrm{y}-2=0$ is

${\displaystyle \frac{|12(-2)-5(3)-2|}{\sqrt{{12}^2+(-5{)}^2}}}={\displaystyle \frac{|-24-15-2|}{13}}={\displaystyle \frac{41}{13}}$

22. Find the equation of a line whose perpendicular distance from the origin is 5 units and angle between the positive direction of the $x$-axis and the perpendicular is ${30}^{\circ }$.

Ans. Given,

Perpendicular distance of the line from the origin is 5 units.

Angle between the positive direction of the $x$-axis and the perpendicular is ${30}^{\circ }$.

Hence,

$p=5,\alpha ={30}^{\circ }$

Required equation is given by $x\cos \alpha +y\sin \alpha =p$.

$x\cos {30}^{\circ }+y\sin {30}^{\circ }=5$

$\sqrt{3}x+y-10=0$

23. Write the equation of the lines for which $\tan \theta ={\displaystyle \frac{1}{2}}$, where $\mathbf{Q}$ is the inclination of the line and $x$ intercept is 4.

Ans. Given,

Slope of the line will become;

$m=\tan \theta ={\displaystyle \frac{1}{2}}$ and 

$x-$intercept, i.e., d = 4

$y={\displaystyle \frac{1}{2}}(x-4)\left[\because y=m(x-d)\right]$

$2y-x+4=0$

24. Find the angle between the $x$-axis and the line joining the points $(3,-1)$ and $(4,-2)$.

Ans: Given,

Two distinct points.

Let $A(3,-1)$ and $B(4,-2)$

Slope of AB,

 $m={\displaystyle \frac{y_2-y_1}{x_2-x_1}}$

$={\displaystyle \frac{-2-(-1)}{4-3}}$

$=-1$

$\because m=\tan \theta$

$\therefore \tan \theta =-1$

$\theta =135$

where  $\theta$ is the angle which AB makes with the positive direction of the x axis.

25. Find the equation of the line intersecting the $x$-axis at a distance of 3 units to the left of origin with slope $-2$.

Ans. Given,

Line intersecting the $x$-axis at a distance of 3 units to the left of origin.

Therefore, the point is $\left(-3,0\right)$

Slope of the line = m $=-2$

Required equation is given by one-point form,

$y-y_1=m\left(x-x_1\right)$

$\Rightarrow y-0=-2(x+3)$

$\Rightarrow 2x+y+6=0$

Short Answer Question (4 Marks)

1. If $p$ is the length of the $\mathrm{\perp }$ from the origin on the line whose intercepts on the axes are a and b. Show that ${\displaystyle \frac{1}{p^2}}={\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{1}{b^2}}$

Ans: Equation of the line is ${\displaystyle \frac{x}{a}}+{\displaystyle \frac{y}{b}}=1$

$\Rightarrow {\displaystyle \frac{x}{a}}+{\displaystyle \frac{y}{b}}-1=0$

The distance of this line from the origin is P

$\therefore P={\displaystyle \frac{\left|{\displaystyle \frac{0}{a}}+{\displaystyle \frac{0}{b}}-1\right|}{\sqrt{{\left({\displaystyle \frac{1}{a}}\right)}^2+{\left({\displaystyle \frac{1}{b}}\right)}^2}}}\left[d={\displaystyle \frac{|ax+by+c|}{\sqrt{a^2+b^2}}}\right]$

${\displaystyle \frac{P}{1}}={\displaystyle \frac{1}{\sqrt{{\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{1}{b^2}}}}}$

${\displaystyle \frac{1}{P}}=\sqrt{{\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{1}{b^2}}}$

Squaring both the sides, 

${\displaystyle \frac{1}{P^2}}={\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{1}{b^2}}$

Hence proved.

2. Find the value of $p$ so that the three lines $3x+y-2=0,px+2y-3=0$ and $2x-y-3=0$ may intersect at one point.

Ans: Given,

Equation of three lines,

$3x+y-2=0.....(i)$

$px+2y-3=0.....(ii)$

$2x-y+3=0.....(iii)$

On solving equation (i) and (iii), 

$x=1$ and $y=-1$

Put x, y in equation (ii), 

$px+2y-3=0$

$p.1+2.\left(-1\right)-3=0$

$p-2-3=0$

$p-5=0$

$p=5$

3. Find the equation to the straight line which passes through the point $(3,4)$ and has an intercept on the axes equal in magnitude but opposite in sign.

Ans: Given the makes equal intercepts in axes but opposite in sign.

Let the intercepts be,

($x-$intercept) a $=a$

($y-$intercept) b $=b$

The equation of the line is given by ${\displaystyle \frac{x}{a}}+{\displaystyle \frac{y}{b}}=1$.

On substituting the values,

${\displaystyle \frac{x}{a}}+{\displaystyle \frac{y}{-a}}=1$

$\Rightarrow x-y=a\dots \dots .(i)$

Since it passes through the point $(3,4)$

$3-4=a$

$a=-1$

Put the value of a in equation (i)

$x-y=-1$

$x-y+1=0$

4. By using the area of $\mathrm{\Delta }.$ Show that the points $(a,b+c),(b,c+a)$ and $(c,a+b)$ are collinear.

Ans: Given,

Three points, let the points be $\text{A}(a,b+c),\text{B}(b,c+a)$ and $\text{C}(c,a+b)$

We know that

Area of triangle is given by 

$\mathrm{\Delta }={\displaystyle \frac{1}{2}}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

Here,

$x_1=a,y_1=b+c$

$x_2=b,y_2=c+a$

$x_3=c,y_3=a+b$

Putting values

$\mathrm{\Delta }={\displaystyle \frac{1}{2}}\left|\begin{array}{ccc}a& b+c& 1\\ b& c+a& 1\\ c& a+b& 1\end{array}\right|$

Applying $C_1\to C_1+C_2$

$\mathrm{\Delta }={\displaystyle \frac{1}{2}}\left|\begin{array}{ccc}a+b+c& b+c& 1\\ b+c+a& c+a& 1\\ c+a+b& a+b& 1\end{array}\right|$

Taking $(a+b+c)$ common from $C_1$

$\mathrm{\Delta }={\displaystyle \frac{1}{2}}(a+b+c)\left|\begin{array}{ccc}1& b+c& 1\\ 1& c+a& 1\\ 1& a+b& 1\end{array}\right|$

Here, $1^{\text{st }}$ and $3^{\text{rd }}$ Column are Identical

Hence value of determinant is zero

$\mathrm{\Delta }={\displaystyle \frac{1}{2}}(a+b+c)\times 0$

$\mathrm{\Delta }=0$

So, $\mathrm{\Delta }=\mathbf{0}$

Hence points $A,B\mathrm{\&}C$ are collinear

5. Find the slope of a line, which passes through the origin, and the midpoint of the line segment joining the point $P(0,-4)$ and $Q(8,0)$

Ans: Given,

Line segment joining the point $P(0,-4)$ and $Q(8,0)$.

Let $M$ be the midpoint of segment PQ then

$M=\left({\displaystyle \frac{x_1+x_2}{2}}+{\displaystyle \frac{y_1+y_2}{2}}\right)$

$=\left({\displaystyle \frac{0+8}{2}},{\displaystyle \frac{-4+0}{2}}\right)$

$=\left(4,-2\right)$

Slope of line joining origin $O\left(0,0\right)$ and the mid-point $M\left(4,-2\right)$,

 $OM={\displaystyle \frac{y_2-y_1}{x_2-x_1}}$

$={\displaystyle \frac{-2-0}{4-0}}$

$={\displaystyle \frac{-1}{2}}$

6. Find the equation of the line passing through the point $(2,2)$ and cutting off intercepts on the axes whose sum is 9.

Ans: Required equation is  ${\displaystyle \frac{x}{a}}+{\displaystyle \frac{y}{b}}=1\dots \dots (i)$

Where,

($x-$intercept) $=a$

($y-$intercept) $=b$

Given,

Sum of intercepts = 9

$\therefore a+b=9$

$b=9-a.....\left(ii\right)$

Substituting the value of equation (ii) in equation (i), 

$\Rightarrow {\displaystyle \frac{x}{a}}+{\displaystyle \frac{y}{9-a}}=1$

This line passes through $(2,2)$

$\therefore {\displaystyle \frac{2}{a}}+{\displaystyle \frac{2}{9-a}}=1$

$a^2-9a+18=0$

$a^2-6a-3a+18=0$

$a(a-6)-3(a-6)=0$

$(a-6)(a-3)=0$

$a=6,3$

$a=6\text{or}a=3$

On substituting the values of a in equation (ii), we get

$b=3\text{or }b=6$

Therefore, the equation of line are,

${\displaystyle \frac{x}{6}}+{\displaystyle \frac{y}{3}}=1{\displaystyle \frac{x}{3}}+{\displaystyle \frac{y}{6}}=1$

7. Reduce the equation $\sqrt{3}x+y-8=0$ into normal form. Find the values $p$ and $\omega$.

Ans: Given,

Equation of a line$\sqrt{3}x+y-8=0$

Let $\sqrt{3}x+y=8\dots \dots .(i)$

$\sqrt{{(\sqrt{3})}^2+{(1)}^2}=2$

Dividing (i) by 2

${\displaystyle \frac{\sqrt{3}}{2}}x+{\displaystyle \frac{y}{2}}=4$

$x\cos {30}^{\circ }+y\cdot \sin {30}^{\circ }=4\dots \dots .(ii)$

On comparing the above equation with the standard from,

$x\cos \omega +y\sin \omega =p$

Where,

$p$is the perpendicular distance from the origin

$\omega$ is the angle between perpendicular and the positive $x-$axis

$p=4$

$\omega ={\displaystyle \frac{\pi }{6}}$

8. Without using the Pythagoras theorem show that the points $(4,4),(3,5)$ and $(-1,-1)$ are the vertices of a right angled $\mathrm{\Delta }$.

Ans: Let given points be $A(4,4),B(3,5)$ and $C(-1,-1)$

Slope of AB,

$m_1={\displaystyle \frac{5-4}{3-4}}=-1$

Slope of BC,

$m_2={\displaystyle \frac{-1-5}{-1-3}}$

$\Rightarrow {\displaystyle \frac{-6}{-4}}={\displaystyle \frac{3}{2}}$

Slope of AC,

$m_3={\displaystyle \frac{-1-4}{-1-4}}=+1$

Since, 

Slope of $AB\times$ slope of $AC=-1$

$m_1\times m_3=-1$

$\Rightarrow AB\mathrm{\perp }AC$                (if the product of slope is $-1$, then the lines are perpendicular)

Hence $\mathrm{\Delta }\text{ABC}$ is right angled at $\text{A}$.

9. The owner of a milk store finds that he can sell 980 litres of milk each week at Rs 14 per litre and 1220 litre of milk each week at Rs 16 per litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17 per litre?

Ans. Let selling price be P along $x$-axis

 demand of milk be D along $y$-axis

We know that the equation of line is

$y=mx+c$

Here, $P$ is along $x$-axis and $D$ is along $y$-axis

So, our equation becomes

$D=mP+c$

Now,

Owner sells 980 litre milk at Rs 14 /litre

So, $D=980\mathrm{\&}P=14$ satisfies the equation

Putting values in (1)

$980=14m+c$

Owner sells 1220 litre milk at Rs $16/$ litre

So, $D=1220\mathrm{\&}P=16$ satisfies the equation

Putting values in (1)

$1220=16m+c$

So, our equations are

$980=14m+c$

$1220=16m+c$

From (A)

$980=14m+c$

$980-14m=c$

Putting value of $\mathrm{c}$ in $(\mathrm{B})$

$1220=16m+980-14m$

$1220-980=16m-14m$

$240=2m$

${\displaystyle \frac{240}{2}}=2$

$120=m$

$m=120$

Putting $m=120$ in $(A)$

$980=14m+c$

$980-14m=c$

$980-14(120)=c$

$980-1680=c$

$-700=c$

$c=-700$

Putting value of $m\mathrm{\&}c$ in (1)

$D=mP+c$

D=120 P-700

Hence, the required equation is $\mathrm{D}=120\mathrm{P}-700$

We need to find how many litres could he sell weekly at Rs $17/$ litre i.e. we need to find $D$ when $P=17$

Putting $P=17$ in the equation

D=120 P-700

D=120(17)-700

D=2040-700

D=1340

Hence when price is Rs $17/$ litre, 1340 litres of milk could be sold.

10. The line through the points $(\text{h},3)$ and $(4,1)$ intersects the line $7x-9y-19=0$ a right angle. Find the value of $h$.

Ans: Given a line joining $(\text{h},3)$ and $(4,1)$

Let $\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)$ be $(\text{h},3)$ and $(4,1)$ respectively.

Find slope by two-point form,

$m_1={\displaystyle \frac{y_2-y_1}{x_2-x_1}}$

 $\Rightarrow {\displaystyle \frac{1-3}{4-h}}={\displaystyle \frac{-2}{4-h}}$

Given line is $7x-9y-19=0$

Slope of the line is given by $m={\displaystyle \frac{-\text{co - efficient of }x}{\text{co - efficient of }y}}$

$m_2={\displaystyle \frac{-7}{-9}}$