Question

The lines y = x - 7 and y = 3x – 19 intersect at the point A. The point B has coordinates (-2, 11).Find the equation of the line that passes through A and B.

Answer 1

Answer:

Equation of the line = 3x + y - 5 = 0

Step-by-step explanation:

Given:

• The lines y = x - 7 and y = 3x - 19 intersect at point A.
• The point B has coordinates (-2, 11)

To Find:

• Equation of the line that passes through A and B

Solution:

By given,

The lines y = x - 7 and y = 3x - 7 intersect.

Therefore solving these two equations, we get the point of intersection.

y = x - 7 ---(1)

y = 3x - 19 ----(2)

Substituting 1 in 2,

x - 7 = 3x - 19

19 - 7 = 3x + x

4x = 12

x = 3

Substitute x in equation 1,

y = 3 - 7

y = -4

Hence the point of intersection is (3, -4) = A

Also by given the line passes through the points A and B. That is, it passes through the points (3, -4) and (-2, 11)

The equation of a line when two points are given is given by,

$\sf \dfrac{y-y_1}{x-x_1} =\dfrac{y_2-y_1}{x_2-x_1}$

Substitute the values,

$\sf \dfrac{y+4}{x-3} =\dfrac{11+4}{-2-3}$

$\sf \dfrac{y+4}{x-3} =\dfrac{15}{-5}$

$\sf \dfrac{y+4}{x-3} =\dfrac{3}{-1}$

Cross multiplying we get,

-y - 4 = 3x - 9

3x + y - 9 + 4 = 0

3x + y - 5 = 0

Therefore the equation of the line is 3x + y - 5 = 0.

Answer 2

Given :-

The lines y = x - 7 and y = 3x – 19 intersect at the point A. The point B has coordinates (-2, 11).

To Find :-

Equation of the line that passes through A and B.

Solution :-

We are given with two equations

$\sf Value \; of \; y\begin{cases}\bf y = x- 7\\ \bf y= 3x-19\end{cases}$

For the equation of line

$\sf \dfrac{y+4}{x-3}=\dfrac{11+4}{-2-(+3)}$

$\sf\dfrac{y+4}{x-3}=\dfrac{11+4}{-2-3}$

$\sf \dfrac{y+4}{x-3} = \dfrac{15}{-5}$

$\sf\dfrac{y+4}{x-3} = \dfrac{-3}{1}$

$\sf y+4=-3(x-3)$

$\sf y+4=-3x+9$

$\sf y + 3x = 9 - 4$

$\sf 3x+y = 5$

$\sf 3x+y-5=0$