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Question (1):
Find the centroid of the triangle whose sides are
x + y - 1 = 0, x - 3y + 3 = 0 and x - y - 1 = 0.
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Answer:
Let AB represent the side x + y - 1 = 0 ...(i)
Let BC represent the side x - 3y + 3 = 0 ... (ii)
Let CA represent the side x - y - 1 = 0 ... (iii)
Solving (i) and (ii) for x and y, we get the coordinates of B, i.e., B(0,1).
Solving (ii) and (iii) , we get the coordinates of C, i.e., C(3,2).
Solving (iii) and (i), we get the coordinates of A, i.e., A(1,0).
The coordinates of the centroid of the triangle are
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Question (2):
If the two vertices of a triangle are (3,-1) and (-2,3) and its orthocentre is the origin, find the coordinates of the third vertex.
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Answer:
In DABC, A(3, -1) and B(-2, 3) and the orthocentre is O(0, 0).
Let AD and BE be the altitudes through A and B.
The third vertex is the intersection of the lines BD and AE.
\ Slope of BD = 3
Solving (i) and (ii) for x and y, we get
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Question (3):
Find the equation of the line which joins the points (a,b) to the
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Answer:
The two given equation are
 The equation of the line through the intersection of (i) and (ii) is (bx + ay - ab) + l(ax + by - ab) = 0.
This line passes through (a,b).
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Question (4):
Prove that following lines are concurrent:
15x - 18y + 1 = 0, 12x + 10y - 3 = 0, 6x + 66y - 11 = 0.
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Answer:
15x - 18y + 1 = 0 ...(i)
12x + 10y - 3 = 0 ...(ii)
6x + 66y - 11 = 0 ...(iii)
Solve (i) and (ii) for x and y,
 0
 RHS
Hence the given lines are concurrent.
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Question (5):
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Answer:
Let the equation of the straight line be y = mx + c or mx - y + c = 0.
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Question (6):
Show that for the line 5x + 6y + 12, (2,-3) and the origin are on the same side.
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Answer:
Let d = distance between the origin and the line.
Let d' = distance between (2,-3) and the line.
Since d and d' are of the same sign, the origin and the point are on the same side of the line.
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Question (7):
Find the equation of the bisectors of the angles between the lines 3x - 4y = 8 and 4x + 3y = 10.
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Answer:
The given equation of lines is rewritten with positive constant.
i.e., -3x + 4y + 8 = 0 and -4x - 3y + 10 = 0
The equation of the bisectors is
Taking only positive values, we get
Taking negative sign only, we get
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Question (8):
The distance of a point (x1, y1) from each of two straight lines, which pass through the origin is d. Prove that the two lines are given by (x1y - xy1)2 = d (x2 + y2).
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Answer:
Let y = mx be any line through the origin. Then, distance d of (x1, y1) 
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Question (9):
Find the equation of the line, which cuts off an intercept of 5 units on negative direction of y-axis and makes an angle of 135o with the positive direction of x-axis.
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Answer:
Here m = 135o = tan (180o - 45o)
= - tan45o
= -1
and c = -5
\ Required equation of the line is
y = (-1)x - 5
y = - x - 5 or x + y + 5 = 0
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Question (10):
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Answer:
 The equation of the line is of the form y = mx + c.
6y  2x - 3
6y - 2x + 3  0
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Question (11):
The vertices of a triangle are (0,0), (a, b) and (b, -a), find the equation to the sides of the triangle.
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Answer:
Let O (0,0), A (a, b) and B (b, -a) be the vertices of the triangle OAB.
 ay = bx or bx - ay = 0
Equation of side OB is
 by = -ax
 ax + by = 0
Equation of the side AB is
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Question (12):
Find the equation of the line through A (3,0) and B (0, -3).
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Answer:
Let P (x, y) be any point on the line joining A and B.
\ Equation of line is
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Question (13):
Reduce each of the following straight line equation to the slope-intercept form. Find m and c.
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Answer:
The slope-intercept form of the equation of line is y = mx + c.
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Question (14):
Find the equation of the line through (1,2) and parallel to x + 3y = 1.
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Answer:
Given line is x + 3y = 1.
The line passes through (1,2).
Let P (x, y) be any point on the required line.
Aliter:
The line parallel to x + 3y - 1 = 0 is x + 3y = k.
This line passes through (1,2).
 Putting x = 1 and y = 2 in x + 3y = k, we have
 The equation of the required line is x + 3y = 7.
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Question (15):
Find the equation of the line with y-intercept 7 and parallel to
4x + 5y = 8.
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Answer:
Reduce the equation to y = mx + c form .
But c = 7,
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