Question

find the distance between the points (4,2) and (4,7) which divides the line externally and internally in the ratio of 2:3

Answer 1

A (4,2) (x1 y1)
B (4,7) (x2 y2)
C (x, y)
m1 = 2, m2 = 3
by sectional formula:
$(x \: y) = ( \frac{m1x2 + m2x1}{m1 + m2} \: \frac{m1y2 + m2y1}{m1 + m2} ) \\ \: = ( \frac{2 \times 4 + 3 \times 4}{2 + 3} \: \frac{2 \times 7 + 3 \times 2}{2 + 3} ) \\ \: = ( \frac{20}{5} \: \frac{20}{5} ) \\ \: = (4 \: 4)$
therefore, C (x, y) = (4, 4)
by distance formula:
AC = √[(4-4)²+(2-4)²]
= √[2]²
= 2 units
BC = √[(4-4)²+(7-4)²]
= √[3]²
= 3 units

therefore the point is (4, 4) which is 2 cm and 3 cm from the two given points.