#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# Prove that the distance from the origin to the orthocentre of the triangle formed by the lines x/m + y/n =1 and ax^2 + 2hxy + by^2 =0 is                (m^2 + n^2)^1/2 {[(a + b) mn] / [am^2 – 2hmn + bn^2]}

Khaled bin seleem
12 Points
one year ago
Let H = (p, q) is orthocenter.
x/m + y/n =1  => nx + my – mn = 0      --------(1)
Opposite vertex to this side is Origin (0, 0).
equation to any line perpendicular to eq (1) and passing through origin is        mx-ny = 0  ------(2)
Now, Length of one altitude OC which is perpendicular to eq (1) is the perpendicular length from origin to eq (1)
i,e.  OC =         $\bg_white \left | -mn \right | / (m^{2}+n^{2})^{1/2}$   =    $\bg_white mn / (m^{2}+n^{2})^{1/2}$               as   $\bg_white \left | a \right | = \pm a$

H lie on this line OC and we have OH = OC – HC
HC is the perpendicular distance from point H to eq (1)
Which is  HC =    $\bg_white \left | np+mq-mn \right | / (m^{2}+n^{2})^{1/2}$        =        $\bg_white (-np-mq+mn) / (m^{2}+n^{2})^{1/2}$     as    $\bg_white \left | a \right | = \pm a$
OH = OC – HC    =      $\bg_white mn / (m^{2}+n^{2})^{1/2}$    –    $\bg_white (-np-mq+mn) / (m^{2}+n^{2})^{1/2}$
=      $\bg_white (np+mq) / (m^{2}+n^{2})^{1/2}$          ----------(3)
Now, let       $ax^{2}+2hxy+by^{2}=0$       contains    y – v1x = 0    -----(4) and y – v2x = 0     --------(5)  as its pair of straight lines.
This implies that        v+ v2 = -2h/b      -------(6)     and     v1v2 = a/b      -------(7)
Solving eq(6) and (7), we get   v1 = $\frac{(h^{2} - ab)^{1/2} - h}{b}$        v2 = $\frac{-((h^{2} - ab)^{1/2} + h)}{b}$

Solving eq (1) and (4) we get the coordinate of another vertex of the triangle. Let us name the vertex and A,
then A =          $(\frac{mn}{n +m v_{1}}, \frac{v_{1}mn}{n +m v_{1}})$

The slope of eq (5) which is the third side of the triangle is    v2 .  the slope of any line perpendicular to (5) is  – 1/v2
the equation to a line with point A and slope -1/v2 is        $x + v_{2}y - \frac{v_{1}v_{2}mn + mn}{n + v_{1}m} = 0$         ---------(8)
eq (2) and (8) represent two altitudes of the given triangle.
Therefore after substituting v1 and v2    values in eq (8) and solving eq (2) and (8)
We get H = (p, q) =         $(\frac{mn^{2} (a + b)}{am^{2} - 2hmn + bn^{2}} , \frac{m^{2}n (a + b)}{am^{2} - 2hmn + bn^{2}} )$

Substituting  p and q values in eq (3) we get
OH =    $(m^{2}+n^{2})^{1/2}(\frac{(a+b)mn}{am^{2}-2hmn+bn^{2}})$

Hence proved.