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?

• A. \( x-7y-12=0 \)
• B. \( x+7y+12=0 \)
• C. \( 7x-y-16=0 \) (This option is \(7x-y-16=0\), image had \(7x+y-16=0\)) The image says \(7x-y-16=0\) with checkmark.
• D. \( 7x+y+16=0 \)

Hint:

In an isosceles right-angled triangle, the equal sides make \(45^\circ\) with the hypotenuse.
Use the formula for angle between two lines: \(\tan\theta = |(m_1-m_2)/(1+m_1m_2)|\).
If slopes are \(m_1, m_2\), they must be perpendicular, so \(m_1 m_2 = -1\).
The equation of a line with slope \(m\) passing through \((x_0, y_0)\) is \(y-y_0 = m(x-x_0)\).
Always check if the options satisfy basic conditions like passing through the given vertex.

Answer 1

The Correct Option is C

Let the isosceles right-angled triangle be ABC, with the right angle at vertex A=(2,2). The hypotenuse BC is given by the line L: \(3x+4y-4=0\). Since the triangle is isosceles and right-angled at A, the sides AB and AC are equal in length and are perpendicular to each other. Also, these sides make an angle of \(45^\circ\) with the hypotenuse BC. The slope of the hypotenuse L is \(m_L = -3/4\). Let the slope of one of the other sides (say AB) be \(m\). The angle \(\theta\) between line L and side AB is \(45^\circ\). Using the formula for the angle between two lines: \(\tan\theta = \left|\frac{m - m_L}{1 + m m_L}\right|\). Here \(\theta = 45^\circ\), so \(\tan 45^\circ = 1\). \[ 1 = \left|\frac{m - (-3/4)}{1 + m(-3/4)}\right| = \left|\frac{m + 3/4}{1 - 3m/4}\right| = \left|\frac{4m+3}{4-3m}\right| \] So, we have two cases: Case 1: \( \frac{4m+3}{4-3m} = 1 \) \( 4m+3 = 4-3m \Rightarrow 7m = 1 \Rightarrow m = 1/7 \). Case 2: \( \frac{4m+3}{4-3m} = -1 \) \( 4m+3 = -(4-3m) = -4+3m \Rightarrow m = -7 \). So the slopes of the two legs AB and AC are \(1/7\) and \(-7\). (Their product is \((1/7)(-7) = -1\), so they are perpendicular, which is correct). Side 1 (slope \(m_1=1/7\)) passes through A(2,2). Equation: \(y-2 = (1/7)(x-2) \Rightarrow 7y-14 = x-2 \Rightarrow x-7y+12=0\). Side 2 (slope \(m_2=-7\)) passes through A(2,2). Equation: \(y-2 = -7(x-2) \Rightarrow y-2 = -7x+14 \Rightarrow 7x+y-16=0\). Let's check the options: % Option (a) \(x-7y-12=0\). My equation is \(x-7y+12=0\). Sign difference. % Option (b) \(x+7y+12=0\). No. % Option (c) \(7x-y-16=0\). My equation is \(7x+y-16=0\). Sign difference on y term. % Option (d) \(7x+y+16=0\). No. The image has checkmark on option (c) which is stated as \(7x-y-16=0\). My calculation for one side was \(7x+y-16=0\). If the equation is \(7x-y-16=0\), then \(y = 7x-16\). Slope is 7. If the slope is 7, then \(\left|\frac{7 - (-3/4)}{1 + 7(-3/4)}\right| = \left|\frac{28/4+3/4}{1 - 21/4}\right| = \left|\frac{31/4}{(4-21)/4}\right| = \left|\frac{31}{-17}\right| = 31/17 \neq 1\). So, slope 7 is incorrect. My slopes \(1/7\) and \(-7\) are correct. The equations derived are \(x-7y+12=0\) and \(7x+y-16=0\). Let's check the options from the image again. Option (c) in the image is \(7x-y-16=0\). If \(7x-y-16=0\), this line passes through A(2,2)? \(7(2) - 2 - 16 = 14 - 2 - 16 = 12 - 16 = -4 \neq 0\). So, option (c) as written in the image does NOT pass through vertex A(2,2). This means there is an error in the question's provided "correct" option or the option text. My derived sides are: Side 1: \(x-7y+12=0\) (Slope 1/7) Side 2: \(7x+y-16=0\) (Slope -7) Option (a) \(x-7y-12=0\). Differs by constant. Option (c) as per image is \(7x-y-16=0\). Slope 7. Does not pass through A(2,2). There is a significant issue with the options matching correct derivation. Assuming my derivation is correct, one of the sides is \(7x+y-16=0\). If one of the options was \(7x+y-16=0\), that would be correct. The option provided as "correct" in the image, \(7x-y-16=0\), does not even pass through the vertex A(2,2). This is a fundamental error. I will list my derived side as the answer. \[ \boxed{\text{One side is } 7x+y-16=0 \text{ or } x-7y+12=0 \text{ (Options provided are problematic)}} \]