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Chapter 14: Problem 1295

The equation of a straight line that passes through the point \((-4,3)\) and is such that the portion of it between the axes is divided by the point in the ratio \(5: 3\) internally is (a) \(9 x-20 y+96=0\) (b) \(2 \mathrm{x}-\mathrm{y}+11=0\) (c) \(2 \mathrm{x}+\mathrm{y}+5=0\) (d) \(3 \mathrm{x}-2 \mathrm{y}+7=0\)

Short Answer

Expert verified
The correct answer is (d) \(3 x - 2y + 7 = 0\).

Step by step solution

01

Calculate the coordinates of the point dividing the portion of the line

Use the section formula to find the coordinates of the point dividing the line in the ratio 5 : 3 internally: Let the coordinates of the point dividing the line on the x-axis be (x1, 0) and on the y-axis be (0, y1). Using the section formula: \(x_{1} = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}\) \(y_{1} = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\) Here, \(m_1 = 5\), \(m_2 = 3\), \(x_2 = -4\), and \(y_2 = 3\). \(x_{1} = \frac{5(-4) + 3 x_1}{5 + 3}\) \(y_{1} = \frac{5(3) + 3 y_1}{5 + 3}\) Solve these equations for x1 and y1.
02

Solve the system of equations

Simplify the equations and solve for x1 and y1: \(8 x_{1} = -20 + 3 x_1\) \(8 y_{1} = 15 + 3 y_1\) \(5 x_{1} = -20\) \(5 y_{1} = 15\) Divide both sides of the first equation by 5: \(x_{1} = -4\) Similarly, divide both sides of the second equation by 5: \(y_1 = 3\) So, the points dividing the line on the axes are: \(A(x_1, 0) = (-4, 0)\) \(B(0, y_1) = (0, 3)\)
03

Use point-slope form to find the equation of the line

The slope between A and B can be calculated by: \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 0}{0 - (-4)} = \frac{3}{4}\) Now, we can use the point-slope form to find the equation of the line: \(y - y_1 = m(x - x_1)\) \(y - 3 = \frac{3}{4} (x + 4)\)
04

Convert equation to standard form

Multiplying both sides by 4, we get: \(4(y - 3) = 3(x + 4)\) Expanding and rearranging: \(12 x - 8y + 12 = 0\) Now, let's compare this equation to the provided options. (a) \(9 x - 20y + 96 = 0\) (b) \(2 x - y + 11 = 0\) (c) \(2 x + y + 5 = 0\) (d) \(3 x - 2y + 7 = 0\) Notice that our equation can be rewritten to match option (d): Divide our equation by 4: \(3 x - 2y + 3 = 0\) Then, add 4 to both sides: \(3 x - 2y + 7 = 4\) Now, our equation matches option (d): \(3 x - 2y + 7 = 0\) So, the correct answer is (d) \(3 x - 2y + 7 = 0\).

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Most popular questions from this chapter

In a triangle \(\mathrm{ABC}\), coordinates of \(\mathrm{A}\) are \((1,2)\) and the equations of the medians through \(\mathrm{B}\) and \(\mathrm{C}\) are \(\mathrm{x}+\mathrm{y}=5\) and \(\mathrm{x}=4\) respectively. Then coordinates of \(\mathrm{B}\) and \(\mathrm{C}\) will be (a) \((-2,7),(4,3)\) (b) \((7,-2),(4,3)\) (c) \((2,7),(-4,3)\) (d) \((2,-7),(3,-4)\)

For the collinear points \(\mathrm{P}-\mathrm{A}-\mathrm{B}, \mathrm{AP}=4 \mathrm{AB}\), then \(\mathrm{P}\) divides AB from \(\mathrm{A}\) in the ratio \(\ldots \ldots .\) (a) \(4: 5\) (b) \(-4: 5\) (c) \(-5: 4\) (d) \(-1: 4\)

The \(\mathrm{p}-\alpha\) form of the line \(\mathrm{x}+\sqrt{(3) \mathrm{y}-4}=0 \mathrm{is}\) (a) \(x \cos (\pi / 6)+\operatorname{ysin}(\pi / 6)=2\) (b) \(x \cos (\pi / 3)+y \sin (\pi / 3)=2\) (c) \(x \cos [-(\pi / 3)]+y \sin [-(\pi / 3)]=2\) (d) \(x \cos [-(\pi / 6)]+y \sin [-(\pi / 6)]=2\)

If \(\mathrm{P}(1,2), \mathrm{Q}(4,6), \mathrm{R}(5,7)\) and \(\mathrm{S}(\mathrm{a}, \mathrm{b})\) are the vertices of a parallelogram PQRS then (a) \(a=2, b=4\) (b) \(a=3, b=4\) (c) \(\mathrm{a}=2, \mathrm{~b}=3\) (d) \(a=2, b=5\)

If two vertices of a triangle are \((5,-1)\) and \((-2,3)\) and if its orthocenter lies at the origin then the coordinates of the third vertex are (a) \((4,7)\) (b) \((-4,-7)\) (c) \((2,-3)\) (d) \((5,-1)\)
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