10th SAMACHEER KALVI SOLUTION FOR EXERCISE 5.5

This page 10th samacheer kalvi solution for exercise 5.5 is going to provide you solution for every problems that you find in the exercise no 5.5

10th samacheer kalvi solution for exercise 5.5

(1) Find the slope of the straight line

(i) 3x + 4y -6 = 0     Solution

(ii) y = 7x + 6     Solution

(iii) 4 x = 5y + 3    Solution

(2) Show that the straight lines x + 2 y + 1 = 0 and 3x + 6y + 2 = 0 are parallel    Solution

(3) Show that the straight lines 3x – 5y + 7 = 0 and 15 x + 9y + 4 = 0 are perpendicular.     Solution

(4) If the straight lines y/2 = x – p and ax + 5 = 3y are parallel,then find a.    Solution

(5) Find the value of a if the straight lines 5x – 2y – 9 = 0 and ay + 2x – 11 = 0 are perpendicular to each other.   Solution

(6) Find the values of p for which the straight lines 8px + (2-3p)y + 1 = 0 and px + 8y + 7 = 0 are perpendicular to each other.   Solution

(7) If the straight lines passing through the points (h,3) and (4,1) intersects the line 7x – 9y – 19 = 0 at right angle,then find the value of h.   Solution

(8) Find the equation of the straight line parallel to the line 3x – y + 7 = 0 and passing through the point (1,-2).   Solution

(9) Find the equation of the straight line perpendicular to the straight line x – 2y + 3 = 0 and passing through the point (1,-2).  Solution

(10) Find the equation of the perpendicular bisector of the straight line segment joining the points (3,4) and (-1,2) 
Solution

(11) Find the equation of the straight line passing through the point of intersection of the lines 2x+y-3=0 and 5x  + y - 6 = 0 and parallel to the line joining the points (1,2) and (2,1) 
Solution

(12) 12. Find the equation of the straight line which passes through the point of intersection of the straight lines 5x - 6y = 1 and 3x + 2y + 5 = 0 and is perpendicular to the straight
line 3x - 5y + 11 = 0.

(13) Find the equation of the straight line joining the point of intersection of the lines 3x – y + 9 = 0 and x + 2y = 4 and the point intersection of the lines 2x + y – 4 = 0 and x – 2y + 3 = 0  Solution

(14)  If the vertices of triangle ABC are A (2,-4) ,B(3,3) and  C (-1,5). Find the equation of the straight line along the altitude from the vertex B.  Solution

(15) If the vertices of triangle ABC are A (-4,4), B(8,4) and C(8,10). Find the equation of the along the median from the vertex A.  Solution

(16) Find the coordinates of the foot of the perpendicular from the origin on the straight line 3x + 2 y = 13   Solution

(17) If x + 2y = 7 and 2 x + y = 8 are the equations of the lines of two diameters of a circle,find the radius of the circle if the point (0,-2) lies on the circle.  Solution

(18) Find the equation of the straight line segment whose end points are the point of intersection of the straight lines 2x – 3y + 4 = 0,x – 2y + 3 = 0 and the midpoint of the line joining the points (3,-2) and (-5,8).

(19) In an isosceles triangle PQR, PQ = PR. The base QR lies on the x axis,P lies on the y-axis and 2x – 3y + 9 = 0 is an equation of PQ. Find the equation of the straight line along PR.

In the page 10th 7th chapter solution for trigonometry part4 we are going to see the solution of next problem





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