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 t

2 thoughts on “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 t”

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,

   [tex]\sf \dfrac{y-y_1}{x-x_1} =\dfrac{y_2-y_1}{x_2-x_1}[/tex]

   Substitute the values,

   [tex]\sf \dfrac{y+4}{x-3} =\dfrac{11+4}{-2-3}[/tex]

   [tex]\sf \dfrac{y+4}{x-3} =\dfrac{15}{-5}[/tex]

   [tex]\sf \dfrac{y+4}{x-3} =\dfrac{3}{-1}[/tex]

   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.

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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

   [tex]\sf Value \; of \; y\begin{cases}\bf y = x- 7\\ \bf y= 3x-19\end{cases}[/tex]

   For the equation of line

   [tex]\sf \dfrac{y+4}{x-3}=\dfrac{11+4}{-2-(+3)}[/tex]

   [tex]\sf\dfrac{y+4}{x-3}=\dfrac{11+4}{-2-3}[/tex]

   [tex]\sf \dfrac{y+4}{x-3} = \dfrac{15}{-5}[/tex]

   [tex]\sf\dfrac{y+4}{x-3} = \dfrac{-3}{1}[/tex]

   [tex]\sf y+4=-3(x-3)[/tex]

   [tex]\sf y+4=-3x+9[/tex]

   [tex]\sf y + 3x = 9 – 4[/tex]

   [tex]\sf 3x+y = 5[/tex]

   [tex]\sf 3x+y-5=0[/tex]

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