Question:
Suppose the hypotenuse and its opposite vertex of an isosceles right angled triangle are \(3x+4y-4=0\) and \((2,2)\) respectively. Then, which of the following is another side of the triangle?
Suppose the hypotenuse and its opposite vertex of an isosceles right angled triangle are \(3x+4y-4=0\) and \((2,2)\) respectively. Then, which of the following is another side of the triangle?
Show Hint
In an isosceles right triangle, the two equal sides make \(45^\circ\) angles with the hypotenuse. Use the formula
\[
\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|
\]
to find the slopes of the equal sides.
Updated On: Jun 26, 2026
- \(x-7y-12=0\)
- \(x+7y+12=0\)
- \(7x+y-16=0\)
- \(7x+y+16=0\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Understand the geometry of the triangle.
In an isosceles right angled triangle, the vertex opposite the hypotenuse is the right angle vertex.
The hypotenuse is given by \[ 3x+4y-4=0 \] and the opposite vertex is \[ (2,2) \] So, the two other sides pass through \((2,2)\).
Since the triangle is isosceles right angled, these two sides are perpendicular to each other and make equal angles with the hypotenuse.
Step 2: Find the slope of the hypotenuse.
The equation of the hypotenuse is \[ 3x+4y-4=0 \] Writing it in slope form, \[ 4y=-3x+4 \] \[ y=-\frac{3}{4}x+1 \] So, the slope of the hypotenuse is \[ m=-\frac{3}{4} \]
Step 3: Use the angle condition.
Let the slope of one side through \((2,2)\) be \(m_1\).
Since the sides make an angle of \(45^\circ\) with the hypotenuse, we use \[ \tan 45^\circ = \left| \frac{m_1-m}{1+mm_1} \right| \] Since \[ \tan 45^\circ=1 \] we get \[ \left| \frac{m_1+\frac{3}{4}}{1-\frac{3m_1}{4}} \right|=1 \] So, \[ \frac{m_1+\frac{3}{4}}{1-\frac{3m_1}{4}}=\pm 1 \]
Step 4: Solve for possible slopes.
First, \[ \frac{m_1+\frac{3}{4}}{1-\frac{3m_1}{4}}=1 \] \[ m_1+\frac{3}{4}=1-\frac{3m_1}{4} \] \[ \frac{7m_1}{4}=\frac{1}{4} \] \[ m_1=\frac{1}{7} \] Second, \[ \frac{m_1+\frac{3}{4}}{1-\frac{3m_1}{4}}=-1 \] \[ m_1+\frac{3}{4}=-1+\frac{3m_1}{4} \] \[ \frac{m_1}{4}=-\frac{7}{4} \] \[ m_1=-7 \] Therefore, the slopes of the two equal sides are \[ \frac{1}{7} \quad \text{and} \quad -7 \]
Step 5: Form the equations of the sides through \((2,2)\).
For slope \(\frac{1}{7}\), \[ y-2=\frac{1}{7}(x-2) \] \[ 7y-14=x-2 \] \[ x-7y+12=0 \] For slope \(-7\), \[ y-2=-7(x-2) \] \[ y-2=-7x+14 \] \[ 7x+y-16=0 \] Among the given options, the matching side is \[ 7x+y-16=0 \]
Step 6: Final conclusion.
Hence, another side of the triangle is \[ \boxed{7x+y-16=0} \] Therefore, the correct option is \[ \boxed{(3)\ 7x+y-16=0} \]
In an isosceles right angled triangle, the vertex opposite the hypotenuse is the right angle vertex.
The hypotenuse is given by \[ 3x+4y-4=0 \] and the opposite vertex is \[ (2,2) \] So, the two other sides pass through \((2,2)\).
Since the triangle is isosceles right angled, these two sides are perpendicular to each other and make equal angles with the hypotenuse.
Step 2: Find the slope of the hypotenuse.
The equation of the hypotenuse is \[ 3x+4y-4=0 \] Writing it in slope form, \[ 4y=-3x+4 \] \[ y=-\frac{3}{4}x+1 \] So, the slope of the hypotenuse is \[ m=-\frac{3}{4} \]
Step 3: Use the angle condition.
Let the slope of one side through \((2,2)\) be \(m_1\).
Since the sides make an angle of \(45^\circ\) with the hypotenuse, we use \[ \tan 45^\circ = \left| \frac{m_1-m}{1+mm_1} \right| \] Since \[ \tan 45^\circ=1 \] we get \[ \left| \frac{m_1+\frac{3}{4}}{1-\frac{3m_1}{4}} \right|=1 \] So, \[ \frac{m_1+\frac{3}{4}}{1-\frac{3m_1}{4}}=\pm 1 \]
Step 4: Solve for possible slopes.
First, \[ \frac{m_1+\frac{3}{4}}{1-\frac{3m_1}{4}}=1 \] \[ m_1+\frac{3}{4}=1-\frac{3m_1}{4} \] \[ \frac{7m_1}{4}=\frac{1}{4} \] \[ m_1=\frac{1}{7} \] Second, \[ \frac{m_1+\frac{3}{4}}{1-\frac{3m_1}{4}}=-1 \] \[ m_1+\frac{3}{4}=-1+\frac{3m_1}{4} \] \[ \frac{m_1}{4}=-\frac{7}{4} \] \[ m_1=-7 \] Therefore, the slopes of the two equal sides are \[ \frac{1}{7} \quad \text{and} \quad -7 \]
Step 5: Form the equations of the sides through \((2,2)\).
For slope \(\frac{1}{7}\), \[ y-2=\frac{1}{7}(x-2) \] \[ 7y-14=x-2 \] \[ x-7y+12=0 \] For slope \(-7\), \[ y-2=-7(x-2) \] \[ y-2=-7x+14 \] \[ 7x+y-16=0 \] Among the given options, the matching side is \[ 7x+y-16=0 \]
Step 6: Final conclusion.
Hence, another side of the triangle is \[ \boxed{7x+y-16=0} \] Therefore, the correct option is \[ \boxed{(3)\ 7x+y-16=0} \]
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