Question

How do you write an equation of a line with slope -4 y-intercept 6 ?

Answer 1

Hint: We know that the standard equation of straight line $y=mx+c$ where m is the slope of the line and c is the y-intercept of the straight line. So if we have the value of slope and y-intercept then we can write the equation $y=mx+c$

Complete step by step answer:
The given slope in the question is -4 and the value of y-intercept is 6
We know that the equation of straight line y = slope $\times$ x + y-intercept
So we can figure out the equation of straight line $y=-4x+6$
Now let’s draw the graph and point out the y-intercept

We can see in the above figure the straight line is $y=-4x+6$ and the y-intercept is point A(0,6)
The equation of the straight line with slope -4 and y-intercept 6 is $y=-4x+6$

Note:
 We can find the equation of the straight line by another method, we know that the y-intercept is 6 so the straight pass through the point (0,6) and the slope of a line is -4. Let the point (x,y) lie on the straight line so the value of slope will be ${\displaystyle \frac{y-6}{x-0}}$ which is equal to -4.
So we can write $-4={\displaystyle \frac{y-6}{x-0}}$
Multiplying x in LHS and RHS we get
$\Rightarrow -4x=y-6$
Now adding 6 in both LHS and RHS we get
$\Rightarrow -4x+6=y$
We can see the above equation is the equation of the straight line that is $y=-4x+6$