Question

1. Find the orthocentre of the triangle (5,1),(1,2), (2,3) as
vertices. (7m)

2. Show that the lines 4x-y+7=0,3x-5y-9=0,x+4y+8=0 form a right
angled isosceles triangle. (6m)

3. Find the points on the line x-y-1=0 which are at distance of v2
units from (1,5). (6m)

4. Find the values of k, if the angle between the straight lines
2x+3y-5=0,5x+ky-6=0 is r/4.(6m)
N​

Answer 1

Given:

Three vertices of a triangle are (5,1) , (1,2) , (2,3)

To find: Orthocentre of a triangle with the given vertices.

Orthocentre: Othocentre in a triangle, is the point of intersection of the altitudes.

Altitude: Altitude is the line which meets an other line perpendicularly, i.e., making an angle of 90°.

Solution:

Let the vertices be denoted as:

 $A (x1,y1) = (5,1)\\B (x2,y2) = (1.2)\\C (x3,y3) = (2,3)$

Now, Slope of the line joining AB can be calculated as:

$= (y2 - y1) / (x2-x1)\\= (2 - 1) / ( 1 - 5)\\= 1/(-4)\\= -1/4$

Slope of the line joining BC can be calculated as:

$= (y3 - y2) / (x3 - x2)\\= (3 - 2) / (2 - 1)\\= 1 / 1\\= 1$

Now, Equation for any line with end co-ordinates and the corresponding slope is given by:

⇒$(y2 - y1) = m (x2 - x1)$

where 'm' is the slope of the line

• Note-1: Slope of two parallel lines will be the same.
• Note-2: If slope of a line is 'm', then the slope of the line perpendicular to it will be $"(-1 / m)"$.

From the figure inserted, let the coordinates of the orthocentre (O) be (x,y).

From the note mentioned above, the equation of the altitude meeting line AB, can be given by:

$(y - y1) = (-1 / m) (x - x1)\\(y - 1) = (-1 / (-1 / 4)) (x - 5)\\(y - 1) = 4 (x - 5)\\y -1 = 4x - 20\\4x - y = 20 - 1\\4x - y = 19.......(1)$

Now, the equation of the altitude meeting line BC, can be given by:

$(y - y2) = (-1 / m) (x - x2)\\(y - 2) = (-1 / 1) (x - 1)\\(y - 2) = (-1) (x - 1)\\(y - 2) = (-x + 1)\\x + y = 3......(2)$

As, we know that orthocentre is the point of intersection of altitudes, we have the equations of two altitudes. Two equations are sufficient to find the solution and that solution is the orthocentre of the triangle.

$4x - y = 19\\x + y = 3$

Adding these two equations, we get:

$5x = 22\\x = 22/5$

Now, to get the value of 'y', put the value of 'x' in equation (2). This gives:

$x + y = 3\\y = 3 - x\\y = 3 - (22 / 5)\\y = (15-22) / 5\\y = -7 / 5$

∴ The orthocentre of the given triangle is $(x , y) = (22/5 , -7/5)$

Hence, found.