Question

Q.2 Find the equation of the straight line passing through the A(-2, -4) and B(1, -7). (3)

Q.3 Find the equation of the straight line passing through (3, 5) and parallel to the line y = 2x+ 3 (2)

Q.4 Find the equation of the straight line passing through (5, 7) and gradient 3

Answer 1

See answers below

Explanation:

Standard equation of a line is expressed as y = mx+b

m is the slope

b is the y intercept

Given the coordinate points (-2, -4) and B(1, -7)

Slope m = y2-y1/x2-x1

m = -7-(-4)/1-(-2)

m = -7+4/1+2

m = -3/3

m = -1

Get the intercept

Substitute m = -1 and (1, -7) into y = mx+b

-7 = -1(1) + b

-7 = -1 + b

b = -7+1

b = -6

Get thw equation

y = mx+b

y = -x + (-6)

y = -x - 6

Q.3 The equation in point slope form is expressed as y - y0 = m(x-x0)

Given the point (3, 5)

x0 = 3

y0 = 5

Given the equation y = 2x+3

mx = 2x

m = 2

slope = 2

Substitute the slope and the point in the expression

y - 5 = 2(x -3)

y - 5 = 2x - 6

y - 2x = -6 + 5

y - 2x = -1

This gives the required equation

Q.4 The equation in point slope form is expressed as y - y0 = m(x-x0)

Given the point (5, 7)

x0 = 5

y0 = 7

Slope = 3

Substitute the slope and the point in the expression

y - 7 = 3(x -5)

y - 7 = 3x - 15

y - 3x = -15 + 7

y - 3x = -8

This give the required equation