Question

hello guys

please answer

the equation of line passing the point of intersection of lines 3x+2y+4=0, 2x+5y-1=0 and whose distance from the point (2,-1) is 2 is

Answer 1

The equation of a line passing through the intersection of 3x+2y+4=0 and 2x+5y-1=0 is given by:

k(3x+2y+4) + 2x+5y-1 = 0

=> (3k+2)x + (2k+5)y + 4k-1 = 0

Now, distance of the above line from (2,-1) is 2.

$\Rightarrow \frac{| (3k+2) \times 2 + (2k+5) \times (-1) + 4k-1 | }{\sqrt{(3k+2)^2 + (2k+5)^2}} = 2 \\ \\ \Rightarrow \frac{| 6k + 4 - 2k - 5 + 4k-1 | }{\sqrt{9 {k}^{2} + 12k + 4 + 4 {k}^{2} + 20k + 25}} = 2 \\ \\ \Rightarrow \frac{| 8k - 2 | }{\sqrt{13 {k}^{2} + 32k + 29}} = 2 \\ \\ \Rightarrow | 8k - 2 | = 2\sqrt{13 {k}^{2} + 32k + 29}$


$\text{Square both sides}\\ \\ \Rightarrow | 8k - 2 | ^2 = \big(2\sqrt{13 {k}^{2} + 32k + 29} \: \big)^2 \\ \\ \Rightarrow 64 {k}^{2} - 32k + 4 = 4(13 {k}^{2} + 32k + 29) \\ \\ \Rightarrow 64 {k}^{2} - 32k + 4 = 52 {k}^{2} + 128k + 116\\ \\ \Rightarrow (64 - 52){k}^{2} -( 32 + 128)k + 4 - 116 = 0 \\ \\ \Rightarrow 12 {k}^{2} - 160k - 112 =0 \\ \\ \Rightarrow 3 {k}^{2} - 40k - 28 =0 \\ \\ \Rightarrow 3 {k}^{2} - 42k + 2k - 28 =0 \\ \\ \Rightarrow 3k(k -14) + 2(k - 14) =0 \\ \\ \Rightarrow (3k + 2)(k -14) = 0 \\ \\ \Rightarrow k = \frac{ - 2}{3} \: \: and \: \: 14$
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$\text{If k = 14, the line is}\\ \\ (3k+2)x + (2k+5)y + 4k-1 = 0\\ \\ \Rightarrow (3 \times 14+2)x + (2 \times 14+5)y + 4 \times 14 -1 = 0\\ \\ \Rightarrow 44x+33y+55=0\\ \\ \Rightarrow 11(4x+3y+5)=0\\ \\ \Rightarrow \boxed{ \bold{\red{ 4x+3y+5=0}}}$


$\text{If k = \frac{ - 2}{3} , the line is}\\ \\ (3k+2)x + (2k+5)y + 4k-1 = 0\\ \\ \Rightarrow (3 \times \frac{ - 2}{3}+2)x + (2 \times \frac{ - 2}{3}+5)y + 4 \times \frac{ - 2}{3} -1 = 0\\ \\ \Rightarrow 0 \times x+\bigg(\frac{ - 4}{3}+5 \bigg) y+\bigg( \frac{-8}{3}-1 \bigg)=0\\ \\ \Rightarrow \frac{11}{3} \: y - \frac{11}{3} = 0 \\ \\ \Rightarrow \frac{11}{3}(y - 1) = 0 \\ \\ \Rightarrow \boxed{ \bold{\red{y - 1 = 0}}}$