First slide

Equation of line in Straight lines

Question

The lines $2 x + 5 y = 10$ and $3 x + 4 y = 12$ cuts the x-axis at A and B respectively. A line L drawn through the point $(3, 4)$ meets y-axis at C such that the abscissae of A ,B, and ordinate of C are in geometric progression. If the equation of the line L is $ax + by + c = 0$ then $a + b + c$ is

Moderate

A

6
B

-37
C

37
D

-27

Solution

The line $2 x + 5 y = 10$ cuts x-axis at $A (5, 0)$ and the line $3 x + 4 y = 12$ cuts x-axis at $B (4, 0)$
The point C lies on y-axis , let it be $C (0, k)$
Given $5, 4, k$ are in geometric progression
It implies that
$\frac{4}{5} = \frac{k}{4}$
Cross multiply
$\begin{matrix} 5 k = 16 \\ k = \frac{16}{5} \end{matrix}$
Hence the line L is passing through the points $(3, 4), (0, \frac{16}{5})$
Equation of L
$\begin{matrix} y - 4 = \frac{\frac{16}{5} - 4}{0 - 3} (x - 3) \\ = \frac{4}{15} (x - 3) \\ 15 y - 60 = 4 x - 12 \\ 4 x - 15 y + 48 = 0 \end{matrix}$
Hence, $a + b + c = 4 - 15 + 48 = 37$

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