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प्रश्न
A(7, −1), B(4, 1) and C(−3, 4) are the vertices of a triangle ABC. Find the equation of a line through the vertex B and the point P in AC; such that AP : CP = 2 : 3.
उत्तर
P divides AC in the ratio of 2 : 3
∴ Co-ordinates of P will be
`((m_1x_2 + m_2x_1)/(m_1 + m_2),(m_1y_2 + m_2y_1)/(m_1 + m_2))`
`((2(-3) + 3(7))/(2 + 3),(2 xx 4 + 3 xx (-1))/(2 + 3))`
= `((-6 + 21)/5, (8 - 3)/5)`
= `(15/5, 5/5)`
= (3, 1)
∴ Slope of line passing through B and P
= `(y_2 - y_1)/(x_2 - x_1)`
= `(1 - 1)/(3 - 4)`
= `0/(-1)`
= 0
∴ Equation of the required line is given by y – y1 = m(x – x1)
`=>` y – 1 = 0(x – 4)
`=>` y – 1 = 0
`=>` y = 1
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