10

Q1. The distance of the line y= -3 from the origin is

• 3
• 1.5 units
• 6 units
• 4 units

Solution

Q2. Find the equation of line passing through the mid - point of line joining the points (3, 4) and (5, 6) and perpendicular to the equation of line 2x + 3y = 5.

• 5x – 7y + 4 = 0
• 2x – 3y – 2 = 0
• 3x – 2y – 2 = 0
• 4x – 5y + 13 = 0

Solution

Consider,  2x + 3y = 5 → 3y = 5 - 2x →  Slope of line 2x + 3y = 5 is . Slope of line perpendicular to the line 2x + 3y = 5 is  Slope of line perpendicular to the line 2x + 3y = 5 is . Mid-point of line joining the points (3, 4) and (5, 6) is (4, 5). Equation of line passing through the point (4, 5) and having slope  is: y – 5 = (x – 4)   2y – 10 = 3x – 12   3x – 2y – 2 = 0

Q3. The distance between the parallel lines  4x-3y+5=0 and 4x-3y+15 =0 is :

• 2 units
• 4 units
• 3.5 units
• 5 units

Solution

Q4. The equation of the line parallel to the line 2x – 3y = 1 and passing through the middle point of the line segment joining the points (1, 3) and (1, –7), is:

• 2x – 3y – 8 = 0
• 2x + 3y – 5 = 0
• 3x – 2y + 8 = 0
• 4x – 6y + 7 = 0

Solution

The middle point of line joining the points (1, 3) and (1, –7) is The slope of line 2x – 3y = 1 is . The slope of line parallel to the line 2x – 3y = 1 is . The equation of line passes through (1, –2) and with slope  is:

Q5. m₁ and m₂ are the slope of two perpendicular lines, if

• m₁ = m₂
• 1 + m₁.m₂ = 0
• m₁.m₂ = 1
• m₁ + m₂ = 0

Solution

Q6. Two lines 3x+4y=8 and   lx+my=n are perpendicular. Which of the following is true?

• 3l+4m=0
• 3l-4m=0
• 3m-4l=0
• 3m+4l=0

Solution

When two lines are perpendicular, the product of their slopes =-1 Here, slopes are -3/4 and –l/m So using the above property, 3l/4m=-1 So, 3l+4m=0

Q7. At what point the origin be shifted, if the coordinates of a point (6,7) becomes (-1,11) ?

• (7,4)
• (7,-4)
• (-7,4)
• (4,-7)

Solution

Q8. The equation of a line with x intercept 4 and y intercept 3 is given by

• x/4+y/3=1
• x/4-y/3=1
• x/3+y/4=1
• x/3-y/4=1

Solution

The equation of a line in the intercept form where the x intercept is a and y intercept is b , is given by x/a+y/b=1. So, substituting a=4 and b=3 we get  the answer.

Q9. The ratio in which the point R (1, 2) divides the line segment joining points P (2, 3) and Q (3, 5) is:

• 2 : 1, internally
• 2 : 1, externally
• 1 : 2, internally
• 1 : 2, externally

Solution

P (2, 3) and Q (3, 5) Let R (1, 2) divides the line segment PQ in the ratio 1 : k. Therefore, the coordinates of point R    2k + 3 = k + 1  2k – k = 1 – 3  k = - 2 Therefore, the point R divides the line segment joining the point P and Q in the ratio 1 : 2 externally.

Q10. Area of the triangle formed by the co-ordiante axes & the line ax + by = 2ab is.

• 2 ab sq. units
• 4 ab sq. units
• 8 ab sq. units
• 16 ab sq. units

Solution

Q11. The coordinates of centroid of triangle whose vertices are A(-1,-3), B(5,-6) and C(2,3) and origin gets shifted to (1,2) :

• (2,-2)
• (1,-4)
• (0,0)
• (4,-1)

Solution

Q12. Slope of a line is not defined, when θ =

• 45°
• 60°
• 30°
• 90°

Solution

  Slope of a line is not defined, when q = 90°, as tan 90° is undefined.

Q13. Find the equation of the line through (2,3) so that the segment of the line intercepted between the axes is bisected at that point.

• 3x - 2y = 12
• 2x - 3y = 12
• 3x + 2y = 12
• 2x + 3y = 12

Solution

Q14. The area of the triangle whose vertices are (2, 3), (3, 5) and (7, 5) is:

• 7 units
• 4 units
• 2 units
• 7/2 units

Solution

Let the vertices of triangle ABC are A (2, 3), B (3, 5) and C (7, 5).     Therefore, the area of triangle ABC is 4 units.

Q15. ____ is the slope or gradient of a line, if q is the inclination of the line with X-axis?

• cos q
• sin q
• tan q
• cot q

Solution

Tan q is the slope or gradient of a line, if q is the inclination of a line.

Q16. The line which makes intercepts 3 and 4 on the x and the y axis respectively, has equation

• 4x+3y=12
• 3x+4y=12
• 4x-3y=12
• 3x-4y=12

Solution

The equation of a line which makes intercepts a and b on the x and y axes respectively, is (x/a)+(y/b)=1. Substituting a=3 and b=4 we get the answer as 4x+3y=12

Q17. What will be the new equation of the straight line 5x + 8y = 10, if the origin gets shifted to (2,-3) ?

• 5x + 8y = 24
• 8x + 5y = 24
• 5x + 8y = -4
• 5x – 8y = 14

Solution

Q18. The coordinate _______ is at 5 units distance from the X-axis measured along the negative Y-axis and has zero distance from the y-axis.

• (0, -5)
• (-5, 0)
• (2, 5)
• (5, 0)

Solution

The coordinate (0, -5) is at 5 units distance from the X-axis measured along the negative y-axis and has zero distance from the Y-axis.

Q19. Find the equation of the line whose intercepts on X and Y-axes are a² and b² respectively

• bx + ay = ab
• b²x + a² b²y = a²
• b²x + a²y = a²b²
• b²a²x + a²y = b²

Solution

Here a = a² and b = b². by intercept form, equation of line is        b²x + a²y = a²b²

Q20. A point P is in the interior of angle BAC, such that P lies on the bisector of angle BAC. What can be said about the distance of PM if PN=2cm where PM and PN are perpendiculars from P on the lines BA and AC?

• PM =2 cm
• PM =3.5 cm
• PM = 2 PN
• PM = 3 PN

Solution

A point on the bisector of an angle is equidistant from the arms of the angle. Hence the answer.

Q21. The line through the point (a, b) and parallel to the line Ax + By + C = 0 is:

• Ax + By + C (a + b) = 0
• Ax + Bb + (a + b) = 0
• Ax + By + k = 0
• A (x – a) + B (y – b) = 0

Solution

Let the line parallel to the line Ax + By + C = 0 is             Ax + By + k = 0                                    (i) It passes through (a, b).         Aa + Bb + k = 0                                   (ii) Subtracting equation (ii) from (i), we get A (x - a) + B (y - b) = 0.

Q22. What will be the new equation of the straight line 3x + 4y = 15, if the origin gets shifted to (1,-3) ?

• 3X+4Y=4
• 4X+3Y=6
• 3X+4Y=24
• 3X-4Y=4

Solution

Q23. Two lines are said to be parallel when the difference of their slopes is

• 1
• 2
• - 1
• 0

Solution

  Two lines are said to be parallel when their slopes are equal. This is the same as saying that the difference of their slopes is zero.

Q24. The points A and B have coordinates (3, 2) and (1, 4) respectively. So, the slope of any line perpendicular to AB is

• 1
• - 1
• 2
• - 2

Solution

Q25. Find the slope of line passing through the point A (5, 4) and B (4, 6).

• 2
• 3
• -4
• -2

Solution

The slope of line through (5, 4) and (4, 6) is m =

Q26. A line is known with its slope and an intercept on one of the axes, then the equation of the line is in the form of:

• Point slope form
• Two point form
• Intercept form
• Slope-intercept form

Solution

A line of the form slope-intercept form is known to with its slope and an intercept on one of the axes.

Q27. Equation of the line through the points (2, 3) and (5, 3) is:

• y = 5
• y = 3
• x – y = 2
• x + 2y = 5

Solution

Equation of the line through the points (2, 3) and (5, 3) is:

Q28. The equation of line passing through the point (3,4) & whose intercept on y-axis is twice that on the x-axis is.

• 2x + y = 10
• 2x - y = 10
• x + 2y = 10
• x - 2y = 10

Solution

Q29. The origin lies on

• x – axis only
• y – axis only
• both axis
• none of the axes

Solution

The origin lies on both x and y axis

Q30. Find the equation of the line parallel to the line 5x – 4y + 3 = 0 and passing through the point (2, 5) is:

• 5x – 4y + 10 = 0.
• 5x – 4y + 20 = 0.
• 4x – 5y + 10 = 0.
• 4x – 5y + 20 = 0.

Solution

Let any line parallel to the line 5x – 4y + 3 = 0 is 5x – 4y = k The line passes through (2, 5)             5 x 2 – 4 x 5 = k Þ        10 – 20 = k Þ        -10 = k Equation of the required line is 5x – 4y = -10 Þ        5x – 4y + 10 = 0.

Q31. Find the intercept cut off by the line 2x - y + 16 = 0 on x-axis.

• -8
• 16
• 8
• -16

Solution

Q32. If the area of a triangle is zero, and then these three points are _______.

• Collinear
• Coincident
• Parallel
• Cannot say

Solution

If the area of a triangle is zero, then these three points are collinear.

Q33. New coordinates of the point (7,-1) would be, if the origin is shifted to the point (1,2) by translation of the axis.

• (6,3)
• (6,-3)
• (-3,6)
• (0,0)

Solution

Let (h, k) be the new coordinates. The origin is shifted to the point (1, 2). Therefore, h=7-1=6 k=-1-2=-3   The new coordinates are (6, -3).

Q34. Considering a fixed distance from the origin, how many points can be plotted on the y axis?

• 3
• 2
• 4
• no points

Solution

There are two such points, one above the origin and the one below it.

Q35. The point where all the angle bisectors of a triangle meet is

• Orthocenter
• Circumcenter
• Incenter
• Excenter

Solution

The point where all angle bisectors meet is called Incenter.

Q36. Equation of line parallel to the line Ax + By + C = 0 is:

• Bx + Ay = C
• x + y = - C
• Ax + By + k = 0
• x + y + k = 0.

Solution

Equation of line parallel to the line Ax + By + C = 0 is Ax + By + k = 0.

Q37. The tangent of the angle which the part of the line above the X-axis makes with the positive direction of the X-axis is

• the slope of a line
• the parallel line
• the perpendicular line
• the concurrent line

Solution

  The slope of a line is the tangent of the angle which the part of the line above the X-axis makes with the positive direction of the X-axis.

Q38. Which of the following lines is parallel to the line with equation 2x+y=3?

• 3x+y=3
• 4x+2y=8
• x+2y=1
• x-y=2

Solution

Two parallel lines will have equal slopes. We see that the equation of the given line can be written as y=-2x+3, So by using slope intercept form, we get, slope of the line=-2. Considering 4x+2y=8 i.e. 2y= -4x+8 i.e. y= -2x+4 So the slope of this line is also -2.  hence this is the line || to the given line.

Q39. The equation of the line parallel to the line 2x – 3y = 1 and passing through the middle point of the line segment joining the points (1, 3) and (1, –7), is:

• 2x – 3y – 8 = 0
• 2x + 3y – 5 = 0
• 3x – 2y + 8 = 0
• 4x – 6y + 7 = 0

Solution

The middle point of line joining the points (1, 3) and (1, –7) is The slope of line 2x – 3y = 1 is . The slope of line parallel to the line 2x – 3y = 1 is . The equation of line passes through (1, –2) and with slope  is:  

Q40. If the slope of the line passing through the points (2, 5) and (x, 1) is 2, then x =

• –2
• 4
• 8
• 0

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