Question

The equations of two equal sides AB and AC of an isosceles ∆ABC are x + y = 5 and 7x – y = 3 respectively. Then the equations of the side BC, if area of ∆ABC = 5 square units, can be
The question has multiple correct options
$
  {\text{A}}{\text{. x - 3y - 1 = 0}} \\
  {\text{B}}{\text{. x - 3y - 21 = 0}} \\
  {\text{C}}{\text{. 3x + y - 2 = 0}} \\
  {\text{D}}{\text{. 3x + y - 12 = 0}} \\
$

Answer 1

Hint – To find all the possible line equations of side BC, we find the line equation of AD the perpendicular bisector from A. Then we find out the tan of the angle between AC and AD, find the area of triangle and equate it to the given. We solve the obtained relations for equations of BC.
Complete Step-by-Step solution:

Given Data, x + y = 5 and 7x – y = 3 are line equations and AB = AC
Let us find the point of intersection of the given lines AB and AC, i.e. the point A
y = 7x – 3, putting this in x + y = 5 we get
x + 7x – 3 = 5 ⟹ 8x = 8 ⟹x = 1
⟹7(1) – y = 3 ⟹y = 4.
Hence A (1, 4).

Now we know the tan function of angle x between two lines is given as Tan x =$|\dfrac{{{{\text{m}}_1} - {{\text{m}}_2}}}{{1 + {{\text{m}}_1}{{\text{m}}_2}}}|$, where ${{\text{m}}_1}$ and ${{\text{m}}_2}$are the slopes of the lines respectively.
Given AB: x + y = 5 ⟹y = -x + 5 -- (slope of a line of the form y = mx + c, is m)
Hence ${{\text{m}}_1}$= -1
And AC: 7x – y = 3 ⟹y = 7x – 3
Hence ${{\text{m}}_2}$= 7
Now Tan A = $|\dfrac{{ - 1 - 7}}{{1 + \left( { - 1} \right)\left( 7 \right)}}|$=$\dfrac{4}{3}$.

Now we consider bisectors of angle A of the isosceles triangle, it is perpendicular and passes through the midpoint of BC. Now the bisectors of the lines AB and AC are given as, \[\dfrac{{{\text{ax + by + c}}}}{{\sqrt {{{\text{a}}^2} + {{\text{b}}^2}} }} = \pm \dfrac{{{\text{ap + bq + c}}}}{{\sqrt {{{\text{a}}^2} + {{\text{b}}^2}} }}\]
Therefore the bisectors of lines AB and AC are, \[\dfrac{{{\text{x + y - 5}}}}{{\sqrt {\text{2}} }} = \pm \dfrac{{{\text{7x - y - 3}}}}{{\sqrt {{7^2} + {1^2}} }}\]
⟹${\text{x + y - 5 = }} \pm \dfrac{{{\text{7x - y - 3}}}}{5}$
We solve for both + and – signs, we get
5x + 5y - 25 = 7x – y – 3 and 5x +5y – 25 = -7x + y + 3
⟹2x – 6y + 22 = 0 and 12x + 4y – 28 = 0
⟹x – 3y + 11 = 0 and 3x + y – 7 = 0
As the slope of the line 3x + y – 7 = 0 is y = -3x – 7, i.e. m = -3 is negative. It is the acute angle bisector of the angle A.
Hence equation of line AD is 3x + y – 7 = 0 and its slope is m = -3.
Therefore slope of line BC is $ - \dfrac{1}{{\text{m}}}$= $ - \dfrac{1}{{ - 3}} = \dfrac{1}{3}$ --- (product of slopes of perpendicular lines = -1)
Hence equation of line BC is y = mx + c ⟹ y = $\dfrac{1}{3}$x + c ⟹ x – 3y + 3c = 0.
Let 3c = k, hence equation of line BC is x – 3y + k = 0.

Now let the angle between lines AB and AC be θ.
Area of a triangle between two lines with an angle θ is given by ∆ = $\dfrac{1}{2}$× AB × AC × Sin θ
Now we know from the figure Tan θ =$\dfrac{4}{3}$, Sin θ = $\dfrac{4}{{\sqrt {{3^2} + {4^2}} }} = \dfrac{4}{5}$and AB = AC (isosceles angled triangle), given area of triangle = 5 square units.
Therefore Area of triangle ∆ = 5 =$\dfrac{1}{2} \times {\text{A}}{{\text{B}}^2} \times \dfrac{4}{3}$
$ \Rightarrow {\text{AB = }}\dfrac{5}{{\sqrt 2 }}$.

The point B lies on the line BC, if the y co-ordinate of the point B is y, the x co-ordinate is
x – 3y + k = 0 ⟹x = 3y – k.
Hence B (3y – k, y).

Now B should also satisfy the line equation of AB, i.e. x + y = 5
⟹3y – k + y = 5
⟹k = 3y + y - 5
Therefore x = 3y – 4y + 5= 5 - y
Hence B (5 - y, y)
We also know A (1, 4) and ${\text{AB = }}\dfrac{5}{{\sqrt 2 }}$
Now using distance between two points formula D = $\sqrt {{{\left( {{{\text{x}}_2} - {{\text{x}}_1}} \right)}^2} + {{\left( {{{\text{y}}_2} - {{\text{y}}_1}} \right)}^2}} $
$
   \Rightarrow {\text{A}}{{\text{B}}^2} = \dfrac{{25}}{2} = {\left( {{\text{5 - y}} - 1} \right)^2} + {\left( {{\text{y}} - 4} \right)^2} \\
   \Rightarrow 2{\left( {{\text{y}} - 4} \right)^2} = \dfrac{{25}}{2} \\
   \Rightarrow \left( {{\text{y - 4}}} \right) = \pm \dfrac{5}{2} \\
$
Hence y =$\dfrac{{13}}{2}{\text{ or }}\dfrac{3}{2}$, i.e. x =$ - \dfrac{3}{2}{\text{ or }}\dfrac{7}{2}$.
We know the slope of BC, hence equation of BC: y = mx + c, i.e.
$
   \Rightarrow \dfrac{{13}}{2} = \dfrac{1}{3} \times - \dfrac{3}{2} + {\text{c}} \\
   \Rightarrow {\text{c = 7}} \\
$ And
$
   \Rightarrow \dfrac{3}{2} = \dfrac{1}{3} \times \dfrac{7}{2} + {\text{c}} \\
   \Rightarrow {\text{c = }}\dfrac{1}{3} \\
$
Hence the possible equations of the BC are as follows:
$
   \Rightarrow {\text{y = }}\dfrac{1}{3}{\text{x + 7 and y = }}\dfrac{1}{3}{\text{x + }}\dfrac{1}{3} \\
   \Rightarrow 3{\text{y - x - 21 = 0 and 3y - x - 1 = 0}} \\
$
Hence x – 3y + 1 = 0 and x – 3y + 21 = 0 are the line equations of BC.

Note – In order to solve questions of this type the key is to apply the concepts of lines and triangles appropriately. It is essential to apply the formula of perpendicular bisectors, formula of tan of angle between two lines and distance between two points. The general form of a line is y = mx + c and sin θ = $\dfrac{{{\text{opposite side}}}}{{{\text{hypotenuse}}}}$.