Question

Find the equation of a straight line which passes through the
point of intersection of 3x + 4y - 1 = 0 and 2x - 5y + 7 = 0 and
which is perpendicular to 4x - 2y + 7 = 0.​

Answer 1

Given lines

• 3x + 4y - 1 = 0⠀⠀----[1]
• 2x - 5y + 7 = 0 ⠀ ----[2]
• 4x - 2y + 7 = 0⠀⠀----[3]

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To find

• Equation of line AB.

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Solution

• Let two points, $\sf{A(x_1 , y_1)}$ and $\sf{B(x , y)}$.

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It is given that two lines i.e., 3x + 4y - 1 = 0 and 2x - 5y + 7 = 0 are intersecting each other.

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⠀⠀❍ Let us find the intersecting point.

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$\: \: \: \: \: \: \: \: \: \:\underline{\sf{\red{Solving\: (1)\: and\: (2),\: we\: get}}}$

• x = -1
• y = 1

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$\bf{\therefore\: The\: intersecting\: point\: is\: A(-1 , 1)}$.

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Now, let's find the slope of the third line.

Writing it in the form of $\rm{y = mx + c}$

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• $\sf{y = 2x + \dfrac{7}{2}}$

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Therefore,

• Slope of the line $(m_1)$ is 2.

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We know that, product of slope of two lines that are perpendicular to each other is -1.

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$\boxed{\bf{\bigstar{m_1 \times Slope\: of\: AB = -1{\bigstar}}}}$

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$\tt:\implies\: \: \: \: \: \: \: \: {2 \times \dfrac{y - 1}{x + 1} = -1}$

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$\tt:\implies\: \: \: \: \: \: \: \: {2(y - 1) = -1(x + 1)}$

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$\tt:\implies\: \: \: \: \: \: \: \: {2y - 2 = -x - 1}$

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$\tt:\implies\: \: \: \: \: \: \: \: {x + 2y = -1 + 2}$

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$\bf:\implies\: \: \: \: \: \: \: \: {x + 2y = 1}$

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Hence,

• The required equation of the line is $\bf{x + 2y = 1}$.