Question

Find the equation of a straight line passing through the point P(−5, 2) and parallel to the line joining the points Q(3, −2) and R(−5, 4)

Answer 1

Slope of the line = `(y_2 - y_1)/(x_2 - x_1)`

Slope of the line QR = `(4 + 2)/(-5 - 3) = 6/(-8) = 3/(-4)`

⇒ `-3/4`

Slope of its parallel = `-3/4`

The given point is p(−5, 2)

Equation of the line is y – y₁ = m(x – x₁)

y – 2 = `-3/4(x + 5)`

4y – 8 = – 3x – 15

3x + 4y – 8 + 15 = 0

3x + 4y + 7 = 0

The equation of the line is 3x + 4y + 7 = 0