Question

Find the equation of a line passing through the point (-3,7) and the point
of intersection of the lines 2x-3y+5 = 0 and 4x+9y = 7.

Answer 1

GIVEN :–

• A line passing through the point (-3,7).

• And Passing from the point of intersection of the lines 2x-3y+5 = 0 and 4x+9y = 7.

TO FIND :–

• Equation of line = ?

SOLUTION :–

• Let's find the point of intersection of the lines 2x-3y+5 = 0 and 4x+9y = 7 –

• First equation –

$\\ \implies\tt 2x-3y+5 = 0$

$\\ \implies\tt 2x + 5 = 3y$

$\\ \implies\tt y =\dfrac{1}{3}(2x + 5) \: \: \: \: \: - - - eq.(1)$

• Put the value of 'y' from eq.(1) in another equation –

$\\ \implies\tt 4x+9 \bigg[\dfrac{1}{3}(2x + 5) \bigg] = 7$

$\\ \implies\tt 4x+3(2x + 5)= 7$

$\\ \implies\tt 4x+6x + 15= 7$

$\\ \implies\tt 10x= 7 - 15$

$\\ \implies\tt 10x= - 8$

$\\ \implies \large{ \boxed{\tt x= - \dfrac{4}{5}}}$

• Put the value of 'x' in eq.(1) –

$\\ \implies\tt y =\dfrac{1}{3} \bigg(2 \times - \dfrac{4}{5} + 5 \bigg)$

$\\ \implies\tt y =\dfrac{1}{3} \bigg( - \dfrac{8}{5} + 5 \bigg)$

$\\ \implies\tt y =\dfrac{1}{3} \bigg(\dfrac{25 - 8}{5}\bigg)$

$\\ \implies \large{ \boxed{\tt y =\dfrac{17}{15}}}$

$\\ \implies\tt Point = \bigg(- \dfrac{4}{5},\dfrac{17}{15} \bigg)$

• We know that Line passing from two points $\tt(x_1 , y_1) , (x_2 , y_2)$ is –

$\\\large \implies{\boxed{\tt y - y_1= \dfrac{y_2-y_1}{x_2-x_1} (x-x_1)}}$

• Hence the required line –

$\\ \implies\tt y - 7 = \dfrac{\dfrac{17}{15} - 7}{- \dfrac{4}{5} + 3} (x + 3)$

$\\ \implies\tt y - 7 = \dfrac{\dfrac{17 - 105}{15} }{\dfrac{ - 4 + 15}{5} } (x + 3)$

$\\ \implies\tt y - 7 = \dfrac{\dfrac{ - 88}{15} }{\dfrac{11}{5} } (x + 3)$

$\\ \implies\tt y - 7 = - \dfrac{8}{3} (x + 3)$

$\\ \implies\tt 3(y - 7)= -8(x + 3)$

$\\ \implies\tt 3y -21= -8x - 24$

$\\\large \implies{\boxed{\tt 8x + 3y + 3 = 0}}$

Answer 2

Refer to attachment !!