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Chapter 14: Problem 1197

The equation of line equidistant from the points \(\mathrm{A}(1,-2)\) and \(\mathrm{B}(3,4)\) and making congruent angles with the coordinate axes is \(\ldots \ldots \ldots\) (a) \(x+y=1\) (b) \(y-x+1=0\) (c) \(\mathrm{y}-\mathrm{x}-1=0\) (d) \(y-x=2\)

Short Answer

Expert verified
The equation of the line equidistant from the points A(1,-2) and B(3,4) and making congruent angles with the coordinate axes is \(y + x = 3\). However, this option is not listed among the given choices.

Step by step solution

01

Find the midpoint of A and B

To find the midpoint (M) of points A(1, -2) and B(3, 4), we use the midpoint formula: M(x, y) = \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\) M(x, y) = \(\left(\frac{1+3}{2}, \frac{-2+4}{2}\right)\) M(x, y) = (2, 1)
02

Use the fact that the slope is -1

As the line makes congruent angles with x and y axes, its slope (m) is -1. An equation for a line can be written in the form: y - y1 = m(x - x1) In this case, we will use the midpoint M(2, 1) as (x1, y1). Plugging in the coordinates and the slope, we get: y - 1 = -1(x - 2)
03

Simplify the equation of the line

Now we simplify the equation obtained in Step 2: y - 1 = -1(x - 2) y - 1 = -x + 2 y + x = 3
04

Compare the equation with the given options

Now we compare the equation of our line (y + x = 3) with the given options: (a) x+y=1 -> This does not match the equation. (b) y-x+1=0 -> Rearranging this, we get y + x = 1. This does not match the equation. (c) y-x-1=0 -> Rearranging this, we get y + x = 1. This does not match the equation. (d) y-x=2 -> Rearranging this, we get y + x = 2(x - 1). Setting x = 1 results in y = -1, so this option also doesn't match the equation. None of the given options match the equation of the line we found (y + x = 3), so the correct answer may not be provided within the options.

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Most popular questions from this chapter

If \(2 \mathrm{x}+2 \mathrm{y}-5=0\) is the equation of the line containing one of the sides of an equilateral triangle and \((1,2)\) is one vertex, then find the equations of the lines containing the other two sides. (a) \(\mathrm{y}=(2+\sqrt{3}) \mathrm{x}-\sqrt{3}, \mathrm{y}=(2+\sqrt{3}) \mathrm{x}+\sqrt{3}\) (b) \(y=(2-\sqrt{3}) x-\sqrt{3}, y=(2+\sqrt{3}) x+\sqrt{3}\) (c) \(y=(2-\sqrt{3}) x+\sqrt{3}, y=(2+\sqrt{3}) x-\sqrt{3}\) (d) \(y=(2+\sqrt{3}) x+\sqrt{3}, y=(2+\sqrt{3}) x-\sqrt{3}\)

The straight line \(7 \mathrm{x}-2 \mathrm{y}+10=0\) and \(7 \mathrm{x}+2 \mathrm{y}-10=0\) forms an isosceles triangle with the line \(\mathrm{y}=2\). Area of the triangle is equal to : (a) \((15 / 7)\) sq unit (b) \((10 / 7)\) sq unit (c) \((18 / 7)\) sq unit (d) \((10 / 13)\) sq unit

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If the \(\mathrm{x}\) -coordinate of the point of intersection of the lines \(3 \mathrm{x}+4 \mathrm{y}=9\) and \(\mathrm{y}=\mathrm{mx}+1\) is an integer, then the integer value of \(\mathrm{m}\) is \(\ldots \ldots\) (a) 2 (b) 0 (c) 4 (d) 1

The line parallel to the \(X\) -axis and passing through the intersection of the lines \(\mathrm{ax}+2 \mathrm{by}+3 \mathrm{~b}=0\) and \(\mathrm{bx}-2 \mathrm{ay}-3 \mathrm{a}=0\) where \((\mathrm{a}, \mathrm{b}) \neq(0,0)\) is: (A) above the X-axis at a distance of \((2 / 3)\) from it (B) above the X-axis at a distance of \((3 / 2)\) from it (C) below the X-axis at a distance of \((2 / 3)\) from it (D) below the X-axis at a distance of \((3 / 2)\) from it
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