Question

A line \(L_1\) passes through the points \( (h,k), (1,2) \) and \( (-3,4) \). The points \( (4,3) \) and \( (h,k) \) lie on the line \(L_2\). Given \(L_1 \perp L_2\), then \( (k-h) \) equals to

• A. 2
• B. \( \frac{1}{2} \)
• C. \( -2 \)
• D. 0

Hint:

For perpendicular lines, the product of slopes is always \( -1 \). Use collinearity first to form one equation, then use perpendicularity to form the second equation.

Answer 1

The Correct Option is C

Step 1: Identify the points on line \(L_1\).
Line \(L_1\) passes through:
\[ (h,k),\quad (1,2),\quad (-3,4). \]
Since these three points lie on the same line, their slopes must be equal.

Step 2: Find the slope of \(L_1\).
Using points \( (1,2) \) and \( (-3,4) \), slope of \(L_1\) is:
\[ m_1=\frac{4-2}{-3-1}. \]
\[ m_1=\frac{2}{-4}=-\frac{1}{2}. \]

Step 3: Use point \( (h,k) \) on \(L_1\).
Since \( (h,k) \) and \( (1,2) \) lie on \(L_1\), their slope is also \( -\frac{1}{2} \):
\[ \frac{k-2}{h-1}=-\frac{1}{2}. \]
Cross-multiplying:
\[ 2(k-2)=-(h-1). \]
\[ 2k-4=-h+1. \]
\[ h+2k=5. \]

Step 4: Use perpendicular condition.
Given \(L_1 \perp L_2\).
Therefore:
\[ m_1m_2=-1. \]
Since \(m_1=-\frac{1}{2}\), we get:
\[ -\frac{1}{2}m_2=-1. \]
\[ m_2=2. \]

Step 5: Use points on line \(L_2\).
Line \(L_2\) passes through \( (4,3) \) and \( (h,k) \).
So its slope is:
\[ m_2=\frac{k-3}{h-4}. \]
Since \(m_2=2\),
\[ \frac{k-3}{h-4}=2. \]

Step 6: Form the second equation.
\[ k-3=2(h-4). \]
\[ k-3=2h-8. \]
\[ k=2h-5. \]

Step 7: Solve both equations.
We have:
\[ h+2k=5 \] and
\[ k=2h-5. \]
Substitute \(k=2h-5\) in \(h+2k=5\):
\[ h+2(2h-5)=5. \]
\[ h+4h-10=5. \]
\[ 5h=15. \]
\[ h=3. \]
Now:
\[ k=2(3)-5=1. \]
Therefore:
\[ k-h=1-3=-2. \]
Final Answer:
\[ \boxed{-2}. \]