Find the third vertex of a triangle, if two of its vertices are at (3, 1) and (0, -2) and the
5x1=5
centroid is at the origin,
10
-S-340
10. If the slope of the line påssing through the points (4, -1) and (6, a) is 3, then find the
value of 'a'. Slope-3 at
11 The co-ordinates of one end of a diameter of a circle are (4,) and the coordinate of the
centre of the circle are (1, -3). Find the coordinates of the other end of diameter.
12. Find the slope of (a cos, 8) and (0, a sin o)
13. Find centroid of vertices of triangle A, B and C are respectively ( 4, 6), (2, -2) and
(2,5)
SECTION - III
Note: 1. Answer all questions.
2. Each question carries 2 marks.
4x2=
14. Show that the points (a, b + c), (b, c + a), (c, a + b) are collinear.
15. If the line passing through (4, -3), (6,0) and the line passing through (a, 7) and
origin are parallel, then find the value of 'a'.
16. Determine the ratio in which the line 3x + y - 9 = 0 divide the segment joining
points (1, 3) and (2, 7):
17. Find trisectional points of line joining (-3,-5) and (-6, -8).
IN TY
Answers
9.
Let the coordinates of the third vertex be (x, y)
Then the centroid of the triangle is at
( x + 3 + 0/3, y + 1 - 2/3 )
i.e., ( x + 3/3, y - 1/3 )
ATQ,
(x + 3)/3 = 0 or, x + 3 = 0, i.e., x = - 3
(y - 1)/3 = 3 or, y - 1 = 9, i.e., y = 10
∴ the coordinates of the third vertex is (- 3, 10)
10.
The given points are (4, - 1) and (6, a)
Then the slope of the line passing through these points be
(a + 1)/(6 - 4) = (a + 1)/2
ATQ,
(a + 1)/2 = 3
or, a + 1 = 6
or, a = 5
Therefore the value of a is 5.
11.
Let the other end of the diameter be (x, y)
Then the centre of the circle is at ( x + 4/2, y - 3/2)
ATQ,
(x + 4)/2 = 1 or, x + 4 = 2, i.e., x = - 2
(y - 3)/2 = - 3 or, y - 3 = - 6, i.e., y = - 3
Therefore the coordinates of the other end of te diameter are (- 2, - 3).
12.
The given points are (a cost, 8) and (0, a sint)
Therefore the slope of the line joining these points be
(a sint - 8)/(0 - a cost)
i.e., (a sint - 8)/(- a cost)
i.e., - tant + 8/a * sect
13.
The given vertices of the triangle are A (4, 6), B (2, - 2) and C (2, 5)
Therefore the coordinates of the centroid of the triangle are
( 4 + 2 + 2/3, 6 - 2 + 5/3 )
i.e., (8/3, 9/3)
i.e., (8/3, 3)
14.
The given points are (a, b + c), (b, c + a) & (c, a + b)
Now the area of the triangle made by these points is
| a b + c 1 |
= 1/2 * | b c + a 1 |
| c a + b 1 |
= 1/2 * [a (c + a - a - b) - (b + c) (b - c) + 1 (ab + b² - c² - ca)] [ expanding along the first row ]
= 1/2 * [ca - ab - b² + c² + ab + b² - c² - ca]
= 1/2 * 0
= 0
Since the area of the triangle made by the given points is zero (0), the given points are collinear.
Thus proved.
Note: We can proceed in another way. We find the equation of the line passing through any two points and then we see if the other point satisfies the equation. If does, the points are collinear; otherwise, not.
15.
The first set of points are (4, - 3) and (6, 0)
So the slope of the line passing through these two points is
(0 + 3)/(6 - 4) = 3/2
The second set of points are (a, 7) and (0, 0)
So the slope of the line passing through these two points is
(0 - 7)/(0 - a) = 7/a
ATQ, those two lines are parallel to each other and thus their slopes will be equivalent.
Then,
3/2 = 7/a
i.e., a = 14/3
16.
The given points are (1, 3) and (2, 7)
So the equation of the line passing through these two points be
(y - 7)/(7 - 3) = (x - 2)/(2 - 1)
or, (y - 7)/4 = (x - 2)/1
or, y - 7 = 4x - 8
or, 4x - y = 1 ..... (1)
The given straight line is
3x + y = 9 ..... (2)
Adding (1) and (2), we get
7x = 10
or, x = 10/7
Then y = 33/7
So the line (2) intersects the line (1) at (10/7, 33/7)
Let (10/7, 33/7) divides the line joining the points (1, 3) and (2, 7) into the ratio m : n
Then,
(2m + n)/(m + n) = 10/7
or, 14m + 7n = 10m + 10n
or, 4m = 3n
or, m/n = 3/4 ..... (3)
(7m + 3n)/(m + n) = 33/7
or, 49m + 21n = 33m + 33n
or, 16m = 12n
or, 4m = 3n
or, m/n = 3/4 ..... (4)
From (3) and (4), we can conclude that the required ratio of intersection is 3 : 4
17.
Points that trisect a line segment joining any two points, divide the line segment into the ratio either 1 : 2 or 2 : 1
The given points are (- 3, - 5) and (- 6, - 8)
When it is 1 : 2, the point of trisection be
( (- 6 - 6)/(1 + 2), (- 10 - 8)/(1 + 2) )
i.e., ( - 12/3, - 18/3 )
i.e., (- 4, - 6)
When it is 2 : 1, the point of trisection be
( (- 3 - 12)/(1 + 2), (- 5 - 16)/(1 + 2) )
i.e., ( -15/3, - 21/3 )
i.e., (- 5, - 7)
∴ the required points are (- 4, - 6) and (- 5, - 7)