Sol : Option D
The area of the triangle having its vertices as P(x1, y1), Q(x2, y2) and R(x3, y3) is given by
1/2 [x1( y2 - y3) + x2( y3 - y1) + x3( y1 - y2)]
= 1/2 [4(12+2) + 10(-2-5) + (-3)(5-12)] = 1/2[4.14 - 10.7 + (-3)(-7)] = 1/2[56 - 70 + 21] = 3.5
Hence, solution is option D
Q.2. Find the equation of straight line passing through (2, 3) and perpendicular to the line 3x + 2y + 4 = 0
A. y=5/3x- 2
B. 3Y=2x+5
C. 3Y=5x-2
D. None of these
Sol : Option B
The given line is 3x + 2y + 4 = 0 or y = -3x / 2 - 2
Any line perpendicular to it will have slope = 2 / 3
Thus equation of line through (2, 3) and slope 2 / 3 is
(y – 3) = 2 / 3 (x – 2)
3y – 9 = 2x – 4
3y – 2x – 5 = 0.
Q.3. Find the coordinates of the point which will divide the line joining the points (3, 5) and (11, 8) externally in the ratio 5: 2.
A. (5/3 , 1/3)
B. (3/49 , 1/10)
C. (49/3 , 10)
D. None of these
Sol : Option C
The external division case will use the formula
x = (mx2 – nx1)/(m – n)
y = (my2 – ny1)/(m – n)
where m:n is 5:2 in our case.
Putting the values you will get points (49/3, 10).
Q.4. Find the coordinate of the point which will divide the line joining the point (2,4) and (7,9) internally in the ratio 1:2?
A. (5/3 , 1/3)
B. (3/8 , 3/11)
C. (8/3 , 11/3)
D. (11/3 , 17/3)
Sol : Option D
The internal division will use the formula (mx2 + nx1)/(m + n)
y = (my2 + ny1)/(m + n).
So, the point becomes (11/3, 17/3).
Q.5. What is the slope of the line passing through the points J (-2, 3) and (2, 7)?
A. 1
B. 2
C. &redic;2
D. 4.&redic;3
Sol : Option A
[(y2 - y1) / (x2 - x1)] = [(7-3) / (2-(-2)) ] = 4/4 = 1
Hence, answer is option A
Q.6. Find the equation of the line whose slope is 3 and y intercept is – 4.
A. y = 2x – 3
B. Y=3x+4
C. Y=3x- 4
D. 4.y=√3x-2
Sol : Option C
Given m = 3 and c = - 4. Substituting values in y = mx + c, we get y = 3x – 4.
Hence, solution is option C
Q.7. Find the equation of the line passing through (2, -1) and parallel to the line 2x – y = 4.
A. y = 2x – 5
B. Y=2x+6
C. Y=√2x +7
D. 4.y=2x+5
Sol : Option A
The given line is 2x – y = 4 ⇒ y = 2x – 4 (Converting into the form of y = mx + c)
Its slope = 2. The slope of the parallel line should also be 2.
Hence for the required line
m = 2 and (x1 , y1) = (2, -1).
Equation = y - y1 /x - x1 = y2 - y1 /x2 - x1
⇒ y - y1 /x - x1 = m
⇒ y – y1 = m (x – x1) ⇒ y – ( - 1) = 2 (x – 2)
⇒ y = 2x – 5.
Q8. Find the equation of the line parallel to the line passing through (5,7) and (2,3) and having x intercept as -4.
Sol : Option C
Slope of the given line = (7 – 3) / (5 – 2) = 4/3.
So the slope of the required line is also 4/3. One point on this line is (-4, 0). Hence the equation of the line is
y – 0 = 4/3 (x + 4) ⇒ 3y = 4x +16.
Q9. Find the coordinates of the circum-centre of the triangle whose vertices are (0, 0), (8,0) and (0,6). Find the Circum-radius also.
A. (4, 3),6
B. (3,4),5
C. (4, 3),5
D. (4, 3),3
Sol : Option B
Circum-centre is the point of intersection of the perpendicular bisectors of the three sides of a triangle.
Let S (x, y) be the circumcentre.
SA = SB = SC. ∴ &redic;(x2 + y2) = &redic; (x-8)2 + (y-2) )2.
&redic;(x2 + y2) = &redic; (x-0)2 + (y-6) )2.
Squaring x2 + y2 = x2 – 16x + 64 + y2. So x = 4.
Also x2 + y2 = x2 + y2 – 12y + 36.
So y = 3. Hence coordinates of the circumcentre is (4, 3).
Circumradius = SA = &redic; x2 + y = &redic;16+9 = 5.
Q10. Find the equation of the line passing through (2, -1) and perpendicular to the line 2x – y = 4.
A. x+2y=0
B. Y=5x-2
C. Y=2x-5
D. None of these
Sol : Option A
Slope of the given line is 2. By comparing with y=mx+c
The perpendicular lines have the product of their slopes= (-1)
So, slope of new line will be (-1/2)
Equation of line will be y=mx+c
M=-1/2
So we get, y= (-1/2)x+k
We are given the point (2,-1) that will satisfy this line.
Putting these coordinates we get k=0 and the line becomes=x+2y=0
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