Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio 3 : 5. Find its co-ordinates of point P.

Also, find the equation of the line through P and parallel to 3x + 5y = 7.

Given points, A (8, 0) and B (16, -8)

By section formula, the co-ordinates of the point P which divides AB in the ratio 3 : 5 is given by

$P = \Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big) \\[1em] = \Big(\dfrac{3 \times 16 + 5 \times 8}{3 + 5}, \dfrac{3 \times -8 + 5 \times 0}{3 + 5}\Big) \\[1em] = \Big(\dfrac{48 + 40}{8}, \dfrac{-24 + 0}{8}\Big) \\[1em] = \Big(\dfrac{88}{8}, \dfrac{-24}{8}\Big) \\[1em] = (11, -3).

Given line equation is,

⇒ 3x + 5y = 7

⇒ 5y = -3x + 7

⇒ y = $-\dfrac{3}{5}x + \dfrac{7}{5}$

Comparing above equation with y = mx + c we get,

Slope = $-\dfrac{3}{5}$

The line parallel to the line 3x + 5y = 7 will have the same slope.

Hence, the slope of the required line = $-\dfrac{3}{5}$

By point-slope form, equation of the required line,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-3) = $-\dfrac{3}{5}$(x - 11)

⇒ 5(y + 3) = -3(x - 11)

⇒ 5y + 15 = -3x + 33

⇒ 3x + 5y = 33 - 15

⇒ 3x + 5y = 18.

Hence, P = (11, -3) and equation of the required line is 3x + 5y = 18.