Question

If (h,k) is the image of the point (3,4) with respect to the line 2x - 3y -5 = 0 and (l,m) is the foot of the perpendicular from (h,k) on the line 3x + 2y + 12 = 0, then lh + mk + 1 = 2x - 3y - 5 = 0.

• A. 5
• B. \(\frac{-1}{34}\)
• C. \(\frac{-3}{34}\)
• D. -3

Answer 1

The Correct Option is A

To find the value of $lh + mk + 1$, where $A = (h, k)$ is the image of point $A = (3, 4)$ with respect to the line $2x - 3y - 5 = 0$, and $(l, m)$ is the foot of the perpendicular from $A'$ to the line $3x + 2y + 12 = 0$, we proceed as follows:

1. Finding the Image $A' = (h, k)$:
Let $A' = (h, k)$ be the image of $A = (3, 4)$ with respect to the line $2x - 3y - 5 = 0$. 
The midpoint $M$ of segment $AA'$ lies on the line. The midpoint is:

$ M = \left( \frac{3+h}{2}, \frac{4+k}{2} \right) $
Since $M$ lies on $2x - 3y - 5 = 0$, we have:

$ 2 \left( \frac{3+h}{2} \right) - 3 \left( \frac{4+k}{2} \right) - 5 = 0 $
Simplify:

$ 3 + h - (6 + \frac{3k}{2}) - 5 = 0 $
$ h - \frac{3k}{2} - 2 = 0 $
Multiply through by 2:

$ 2h - 3k = 4 $
Additionally, line $AA'$ is perpendicular to the line $2x - 3y - 5 = 0$, which has slope $\frac{2}{3}$. Thus, the slope of $AA'$ is $-\frac{3}{2}$. 
The slope of $AA'$ is:

$ \frac{k-4}{h-3} = -\frac{3}{2} $
Cross-multiply:

$ 2(k-4) = -3(h-3) $
$ 2k - 8 = -3h + 9 $
$ 3h + 2k = 17 $
Solve the system:

$ 2h - 3k = 4 \quad (1) $
$ 3h + 2k = 17 \quad (2) $
Multiply (1) by 2 and (2) by 3 to eliminate $k$:

$ 4h - 6k = 8 $
$ 9h + 6k = 51 $
Add the equations:

$ 13h = 59 $
$ h = \frac{59}{13} $
Substitute $h$ into (2):

$ 3 \left( \frac{59}{13} \right) + 2k = 17 $
$ \frac{177}{13} + 2k = 17 $
$ 2k = 17 - \frac{177}{13} = \frac{221 - 177}{13} = \frac{44}{13} $
$ k = \frac{22}{13} $
Thus, $A' = \left( \frac{59}{13}, \frac{22}{13} \right)$. 

2. Finding the Foot of the Perpendicular $(l, m)$:
Let $(l, m)$ be the foot of the perpendicular from $A' = \left( \frac{83}{13}, -\frac{14}{13} \right)$ to the line $3x + 2y + 12 = 0$. 
The slope of the line $3x + 2y + 12 = 0$ is $-\frac{3}{2}$, so the perpendicular slope is $\frac{2}{3}$. 
The slope of the line from $A'$ to $(l, m)$ is:

$ \frac{m + \frac{14}{13}}{l - \frac{83}{13}} = \frac{2}{3} $
Cross-multiply:

$ 3 \left( m + \frac{14}{13} \right) = 2 \left( l - \frac{83}{13} \right) $
$ 3m + \frac{42}{13} = 2l - \frac{166}{13} $
Multiply through by 13:

$ 39m + 42 = 26l - 166 $
$ 26l - 39m = 208 \quad (1) $
Since $(l, m)$ lies on $3x + 2y + 12 = 0$, we have:

$ 3l + 2m + 12 = 0 $
$ 3l + 2m = -12 \quad (2) $
Solve the system. Multiply (1) by 2 and (2) by 39 to eliminate $m$:

$ 52l - 78m = 416 $
$ 117l + 78m = -468 $
Add the equations:

$ 169l = -52 $
$ l = -\frac{52}{169} = -\frac{4}{13} $
Substitute $l$ into (2):

$ 3 \left( -\frac{4}{13} \right) + 2m = -12 $
$ -\frac{12}{13} + 2m = -12 $
$ 2m = -12 + \frac{12}{13} = \frac{-156 + 12}{13} = -\frac{144}{13} $
$ m = -\frac{72}{13} $
Thus, the foot of the perpendicular is $\left( -\frac{4}{13}, -\frac{72}{13} \right)$.

3. Calculating $lh + mk + 1$:
Using $h = \frac{83}{13}$, $k = -\frac{14}{13}$, $l = -\frac{4}{13}$, $m = -\frac{72}{13}$, compute:

$ lh + mk + 1 = \left( -\frac{4}{13} \right) \left( \frac{83}{13} \right) + \left( -\frac{72}{13} \right) \left( -\frac{14}{13} \right) + 1 $
$ = \frac{-332}{169} + \frac{1008}{169} + \frac{169}{169} $
$ = \frac{-332 + 1008 + 169}{169} = \frac{845}{169} = 5 $

Final Answer:
The value of $lh + mk + 1$ is $5$.