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Chapter 14: Problem 1330

Let \(\mathrm{A}(\mathrm{h}, \mathrm{k}), \mathrm{B}(1,1)\) and \(\mathrm{C}(2,1)\) be the vertices of right angled triangle with \(\underline{\mathrm{AC}}\) as its hypotenuse. If the area of a triangle is 1 , then the set of \(\mathrm{v}\) values which ' \(\mathrm{k}\) ' can take is given by (a) \((1,3)\) (b) \((0,2)\) (c) \((-1,3)\) (d) \((-3,-2)\)

Short Answer

Expert verified
The set of values that k can take is (-1, 3). Hence, the correct option is (c) (-1, 3).

Step by step solution

01

Calculate the Lengths of the Sides using the Distance Formula

To find the lengths of AB, AC, and BC, we will use the distance formula: \( \) Distance \(= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) Applying this formula, we have: \( \) Length AB \(=\sqrt{(h - 1)^2 + (k - 1)^2} \) \( \) Length AC \(=\sqrt{(h - 2)^2 + (k - 1)^2} \) \( \) Length BC \(= \sqrt{(2 - 1)^2 + (1 - 1)^2} = 1 \)
02

Apply the Pythagorean Theorem

The triangle is right-angled with AC as its hypotenuse. So, applying Pythagorean theorem, we get: (Length AB)^2 + (Length BC)^2 = (Length AC)^2 \((\sqrt{(h - 1)^2 + (k - 1)^2})^2 +(1)^2 =(\sqrt{(h - 2)^2 + (k - 1)^2})^2\)
03

Use the Area of Triangle to Find the Values of k

Since the area of the triangle is 1 and BC has a length of 1, the height from vertex A to BC can be calculated as: Area = (1/2) * Base * Height 1 = (1/2) * 1 * Height Height = 2 We know that the height of the triangle is the absolute difference between the y-coordinates of points A and B, which is |k - 1|. So we have: |k - 1| = 2 Solving for k, we get two possible values: k - 1 = 2 k = 3 Or, k - 1 = -2 k = -1 So the set of values k can take is (-1, 3). Hence, the correct option is (c) (-1, 3).

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Most popular questions from this chapter

One side of the rectangle lies along the line \(4 \mathrm{x}+7 \mathrm{y}+5=0\). Two of its vertices are \((-3,1)\) and \((1,1)\). Then the equations of other side is (a) \(7 x-4 y+25=0\) (b) \(4 x+7 y=11\) (c) \(7 x-4 y-3=0\) (d) All of these

Find the equation of line passing through the point \((\sqrt{3},-1)\) and at a distance \(\sqrt{2}\) units from the origin. (a) \((\sqrt{3}+1) \mathrm{x}+(\sqrt{3}-1) \mathrm{y}=4\) or \((\sqrt{3}-1) \mathrm{x}-(\sqrt{3}+1) \mathrm{y}=4\) (b) \((\sqrt{3}+1) \mathrm{x}+(\sqrt{3}+1) \mathrm{y}=4\) or \((\sqrt{3}-1) \mathrm{x}+(\sqrt{3}+1) \mathrm{y}=4\) (c) \((\sqrt{3}+1) \mathrm{x}+(\sqrt{3}-1) \mathrm{y}=4\) or \((\sqrt{3}-1) \mathrm{x}+(\sqrt{3}+1) \mathrm{y}=4\) (d) \((\sqrt{3}-1) \mathrm{x}+(\sqrt{3}-1) \mathrm{y}=4\) or \((\sqrt{3}+1) \mathrm{x}+(\sqrt{3}+1) \mathrm{y}=4\)

If \(2 \mathrm{x}+3 \mathrm{y}=8\) is perpendicular to the line \((\mathrm{x}+\mathrm{y}+1)+\lambda(2 \mathrm{x}-\mathrm{y}-1)=0\), then \(\lambda=?\) (a) \(-5\) (b) \((3 / 2)\) (c) 5 (d) 0

The nearest point on the line \(3 \mathrm{x}+4 \mathrm{y}=1\) from origin is (a) \([(7 / 25),(4 / 25)]\) (b) \([(7 / 25),(2 / 25)]\) (c) \([(3 / 25),(4 / 25)]\) (d) \([(1 / 25),(3 / 25)]\)

The equation of the lines with slope \(-2\) and intersecting \(\mathrm{x}\) -axis at points distance 3 unit from the origin is...... (a) \(2 \mathrm{x}+\mathrm{y} \pm 6=0\) (b) \(x+2 y \pm 6=0\) (c) \(2 \mathrm{x}+\mathrm{y} \pm 3=0\) (d) \(x+2 y \pm 3=0\)
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