**Exercise 12.1**

**1. Find the slope of a line whose inclination is**

**(i) 45°**

**(ii) 30°**

**Solution:**

**2. Find the inclination of a line whose gradient is**

**(i) 1**

**(ii) √3**

**(iii) 1/√3**

**Solution:**

**5. Find the equation of a straight line parallel to y-axis and passing through the point ( – 3,5).**

**Solution:**

**6. Find the equation of a line whose**

**(i) slope = 3,y-intercept = – 5**

**(ii) slope = -2/7,y-intercept = 3**

**(iii) gradient = √3,y-intercept = -4/3**

**(iv) inclination = 30°,y-intercept = 2**

**Solution:**

**7. Find the slope and y-intercept of the following lines:**

**(i) x – 2y – 1 = 0**

**(ii) 4x – 5y – 9 = 0**

**(iii) 3x + 5y + 7 = 0**

**(iv) x/3 + y/4 = 1**

**(v) y – 3 = 0**

**(vi) x – 3 = 0**

**Solution:**

**8. The equation of the line PQ is 3y – 3x + 7 = 0**

**(i) Write down the slope of the line PQ.**

**(ii) Calculate the angle that the line PQ makes with the positive direction of x-axis.**

**Solution:**

**9. The given figure represents the line y = x + 1 and y = √3x – 1. Write down the angles which the lines make with the positive direction of the x-axis. Hence determine θ.**

**Solution:**

**10. Find the value of p, given that the line y/2 = x – p passes through the point (– 4,4)**

**Solution:**

**11. Given that (a,2a) lies on the line y/2 = 3x – 6. Find the value of a.**

**Solution:**

**12. The graph of the equation y = mx + c passes through the points (1,4) and (– 2,– 5). Determine the values of m and c.**

**Solution:**

**13. Find the equation of the line passing through the point (2,– 5) and making an intercept of – 3 on the y-axis.**

**Solution:**

**14. Find the equation of a straight line passing through (– 1,2) and whose slope is 2/5.**

**Solution:**

_{1}= m (x – x

_{1})

_{1},y

_{1}) is (-1,2)

**15. Find the equation of a straight line whose inclination is 60° and which passes through the point (0,– 3).**

**Solution:**

_{1},y

_{1})

**18. Find the intercepts made by the line 2x – 3y + 12 = 0 on the co-ordinate axis.**

**Solution:**

**19. Find the equation of the line passing through the points P (5,1) and Q (1,– 1). Hence, show that the points P,Q and R (11,4) are collinear.**

**Solution:**

_{2}– y

_{1}/ x

_{2}– x

_{1}

_{1}= m (x – x

_{1})

**20. Find the value of ‘a’ for which the following points A (a,3),B (2,1) and C (5,a) are collinear. Hence find the equation of the line.**

**Solution:**

**21. Use a graph paper for this question. The graph of a linear equation in x and y, passes through A (– 1,– 1) and B (2,5). From your graph, find the values of h and k, if the line passes through (h,4) and (½,k).**

**Solution:**

**23. In ∆ABC,A (3,5),B (7,8) and C (1,– 10). Find the equation of the median through A.**

**Solution:**

_{2 }– y

_{1}/ x

_{2}– x

_{1}

_{1}= m (x – x

_{1})

**24. Find the equation of a line passing through the point (– 2,3) and having x-intercept 4 units.**

**Solution:**

**25. Find the equation of the line whose x-intercept is 6 and y-intercept is – 4.**

**Solution:**

_{2}– y

_{1}/ x

_{2}– x

_{1}

_{1}= m (x – x

_{1})

**27. Find the equation of the line passing through the point (1,4) and intersecting the line x – 2y – 11 = 0 on the y-axis.**

**Solution:**

_{2}– y

_{1 }/ x

_{2}– x

_{1}

_{2}– y

_{1}/ x

_{2}– x

_{1}

_{1 }= m (x – x

_{1})

**30. A and B are two points on the x-axis and y-axis respectively. P (2,– 3) is the mid point of AB. Find the**

**(i) the co-ordinates of A and B.**

**(ii) the slope of the line AB.**

**(iii) the equation of the line AB.**

**Solution:**

_{2}– y

_{1}/ x

_{2}– x

_{1}

_{1}= m (x – x

_{1})

**33. Point A (3,– 2) on reflection in the x-axis is mapped as A’ and point B on reflection in the y-axis is mapped onto B’ ( – 4,3).**

**(i) Write down the co-ordinates of A’ and B.**

**(ii) Find the slope of the line A’B, hence find its inclination.**

**Solution:**

_{2}– y

_{1}/ x

_{2}– x

_{1}

**Exercise 12.2**

**1. State which one of the following is true: The straight lines y = 3x – 5 and 2y = 4x + 7 are**

**(i) parallel**

**(ii) perpendicular**

**(iii) neither parallel nor perpendicular.**

**Solution:**

We have given that:

Given straight lines: y = 3x – 5 and 2y = 4x + 7 ⇒ y = 2x + 7/2

And, their slopes are 3 and 2

The product of slopes is 3 x 2 = 6.

Hence, as the slopes of both the lines are neither equal nor their product is -1 the given pair of straight lines are neither parallel nor perpendicular.

**2. If 6x + 5y – 7 = 0 and 2px + 5y + 1 = 0 are parallel lines, find the value of p.**

**Solution:**

We have given that:

For two lines to be parallel, their slopes must be same.

Given line equations,

6x + 5y – 7 = 0 and 2px + 5y + 1 = 0

In equation 6x + 5y – 7 = 0,

5y = -6x + 7

y = (-6/5) x + 7/5

So, the slope of the line (m_{1}) = -6/5

Again, in equation 2px + 5y + 1 = 0

5y = -2px – 1

y = (-2p/5) x – 1/5

So, the slope of the line (m_{2}) = -2p/5

For these two lines to be parallel

m_{1} = m_{2}

-6/5 = -2p/5

p = (-6/5) x (-5/2)

Thus, p = 3

Hence, the given values p = 3 are not roots of the equation.

**3. Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel. Find the relation connecting a and b.**

**Solution:**

We have given that:

Given lines are: 2x – by + 5 = 0 and ax + 3y = 2

If two lines to be parallel then their slopes must be equal.

In equation 2x – by + 5 = 0,

by = 2x + 5

y = (2/b) x + 5/b

So, the slope of the line (m_{1}) = 2/b

And in equation ax + 3y = 2,

3y = -ax + 2

y = (-a/3) x + 2/3

So, the slope of the line (m_{2}) = (-a/3)

As the lines are parallel

m_{1} = m_{2}

2/b = -a/3

6 = -ab

Hence, the relation connecting a and b is ab + 6 = 0

**4. Given that the line y/2 = x – p and the line ax + 5 = 3y are parallel, find the value of a.**

**Solution:**

We have given that:

Line equation: y/2 = x – p

⇒ y = 2x – 2p

Here, the slope of the line is 2.

And, another line equation: ax + 5 = 3y

⇒ 3y = ax + 5

y = (a/3) x + 5/3

Hence, the slope of the line is a/3

As the line are parallel, their slopes must be equal

⇒ 2 = a/3

a = 6

Hence, the given values a = 6 are not roots of the equation.

Thus, the value of a is 6.

**5. If the lines y = 3x + 7 and 2y + px = 3 perpendicular to each other, find the value of p.**

**Solution:**

We have given that:

If two lines are perpendicular, then the product of their slopes is -1

Now, slope of the line y = 3x + 7 is m_{1 } = 3

And,

The slope of the line: 2y + px = 3

2y = -px + 3

y = (-p/2) x + 3

m_{2 } = -p/2

As the lines as perpendicular,

⇒ m_{1}× m_{2}= -1

3 × (-p/2) = -1

p = 2/3

Hence, the given valuesp = 2/3 are not roots of the equation.

Thus, the value of p is 2/3.

**6. If the straight lines kx – 5y + 4 = 0 and 4x – 2y + 5 = 0 are perpendicular to each other. Find the value of k.**

**Solution:**

_{1}) = k/5

_{2}) = 2

_{1}× m

_{2}= -1

**7. If the lines 3x + by + 5 = 0 and ax – 5y + 7 = 0 are perpendicular to each other, find the relation connecting a and b.**

**Solution:**

_{1}) = -3/b

_{2}) = a/5

_{1}× m

_{2}= -1

**8. Is the line through (– 2,3) and (4,1) perpendicular to the line 3x = y + 1?**

**Does the line 3x = y + 1 bisect the join of (– 2,3) and (4,1).**

**Solution:**

_{1}= y

_{2}– y

_{1}/ x

_{2}– x

_{1}

_{2}) = 3

_{1}× m

_{2}= -1/3 × 3 = -1

**9. The line through A (– 2,3) and B (4,b) is perpendicular to the line 2x – 4y = 5. Find the value of b.**

**Solution:**

_{1}= (b – 3)/ (4 + 2) = (b – 3)/ 6

_{2}= ½

_{1}× m

_{2}= -1

**10. If the lines 3x + y = 4,x – ay + 7 = 0 and bx + 2y + 5 = 0 form three consecutive sides of a rectangle, find the value of a and b.**

**Solution:**

_{1}) = -3

_{2}) = 1/a

_{3}) = -b/2

_{1}× m

_{2}= -1 and m

_{2}× m

_{3}= -1

**11. Find the equation of a line, which has the y-intercept 4, and is parallel to the line 2x – 3y – 7 = 0. Find the coordinates of the point where it cuts the x-axis.**

**Solution:**

**12. Find the equation of a straight line perpendicular to the line 2x + 5y + 7 = 0 and with y-intercept – 3 units.**

**Solution:**

**15. Find the equation of the line passing through (0,4) and parallel to the line 3x + 5y + 15 = 0.**

**Solution:**

_{1 }= m (x – x

_{1})

**16. The equation of a line is y = 3x – 5. Write down the slope of this line and the intercept made by it on the y-axis. Hence or otherwise, write down the equation of a line which is parallel to the line and which passes through the point (0,5).**

**Solution:**

_{1}) = 3

_{1}= m (x – x

_{1})

**17. Write down the equation of the line perpendicular to 3x + 8y = 12 and passing through the point (– 1,– 2).**

**Solution:**

_{1}) = -3/8

_{2}

_{1}× m

_{2}= -1

_{2}= -1

^{2}= 8/3

_{1}= 4/3

_{2}

_{1}× m

_{2}= -1

_{2}= -1

_{2}= -3/4

**19. Find the equation of the line that is parallel to 2x + 5y – 7 = 0 and passes through the mid-point of the line segment joining the points (2,7) and (– 4,1).**

**Solution:**

_{1}= m (x – x

_{1})

**20. Find the equation of the line that is perpendicular to 3x + 2y – 8 = 0 and passes through the mid-point of the line segment joining the points (5,– 2) and (2,2).**

**Solution:**

_{1}) = -3/2

_{2}

_{1}× m

_{2}= -1

_{2}= -1

_{2}= 2/3

_{2}and passing through (7/2,0) will be

**21. Find the equation of a straight line passing through the intersection of 2x + 5y – 4 = 0 with x-axis and parallel to the line 3x – 7y + 8 = 0.**

**Solution:**

**22. The equation of a line is 3x + 4y – 7 = 0. Find (i) the slope of the line. (ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0.**

**Solution:**

_{1}) = -3/4

_{2}

_{1}× m

_{2}= -1

_{2 }= -1

_{2 }= 4/3

**23. Find the equation of the line perpendicular from the point (1,– 2) on the line 4x – 3y – 5 = 0. Also find the co-ordinates of the foot of perpendicular.**

**Solution:**

_{1}) = 4/3

_{2}

_{1}× m

_{2}= -1

_{2 }= -1

_{2 }= -3/4

**24. Prove that the line through (0,0) and (2,3) is parallel to the line through (2,– 2) and (6,4).**

**Solution:**

_{1}

_{1}= (y

_{2}– y

_{1})/ (x

_{2}– x

_{1})

_{2}

_{2 }= (y

_{2}– y

_{1})/ (x

_{2}– x

_{1})

_{1}= m

_{2}

**25. Prove that the line through (– 2,6) and (4,8) is perpendicular to the line through (8,12) and (4,24).**

**Solution:**

_{1}

_{1}= (y

_{2}– y

_{1})/ (x

_{2}– x

_{1})

_{2}

_{2 }= (y

_{2}– y

_{1})/ (x

_{2}– x

_{1})

_{1}× m

_{2}= 1/3 x (-3) = -1

**26. Show that the triangle formed by the points A (1,3),B (3,– 1) and C (– 5,– 5) is a right-angled triangle by using slopes.**

**Solution:**

_{1}= (-1 – 3)/ (3 – 1) = -4/2 = -2

_{2}= (-5 + 1)/ (-5 – 3) = -4/-8 = ½

_{1}× m

_{2}= (-2) x (1/2) = -1

**27. Find the equation of the line through the point (– 1,3) and parallel to the line joining the points (0,– 2) and (4,5).**

**Solution:**

_{1}= m (x – x

_{1}) ⇒ y – 3 = 7/4 (x + 1)

**28. Are the vertices of a triangle.**

**(i) Find the coordinates of the centroid G of the triangle.**

**(ii) Find the equation of the line through G and parallel to AC.**

**Solution:**

_{1}+ x

_{2 }+ x

_{3})/2,(y

_{1}+ y

_{2}+ y

_{3})/2)

_{2}– y

_{1})/ (x

_{2}– x

_{1}) = (-2 – 3)/ (3 – (-1)) = -5/4

**29. The line through P (5,3) intersects y-axis at Q. (i) Write the slope of the line. (ii) Write the equation of the line. (iii) Find the coordinates of Q.**

**Solution:**

_{2}– y

_{1}/ x

_{2}– x

_{1}

**30. In the adjoining diagram, write down (i) the co-ordinates of the points A,B and C. (ii) the equation of the line through A parallel to BC.**

**Solution:**

**31. Find the equation of the line through (0,– 3) and perpendicular to the line joining the points (– 3,2) and (9,1).**

**Solution:**

_{2}= (1 – 2)/ (9 + 3) = -1/12

_{2}

_{1}× m

_{2}= -1

_{2 }= -1

_{2 }= 12

_{2 }will be

**32. The vertices of a triangle are A (10,4),B (4,– 9) and C (– 2,– 1). Find the equation of the altitude through A. The perpendicular drawn from a vertex of a triangle to the opposite side is called altitude.**

**Solution:**

_{2}

_{1 }× m

_{2 }= -1

_{2 }= -1

_{2 }= ¾

**33. A (2, – 4), B (3, 3) and C (– 1, 5) are the vertices of triangle ABC. Find the equation of:**

**(i) the median of the triangle through A**

**(ii) the altitude of the triangle through B.**

**Solution:**

_{1}) = (5 + 4)/ (-1 – 2) = 9/-3 = -3

_{2}

_{1 }× m

_{2}= -1

_{2}= -1

_{2 }= 1/3

**34. Find the equation of the right bisector of the line segment joining the points (1,2) and (5,– 6).**

**Solution:**

_{2 }is the slope of the right bisector of the above line

_{1 }× m

_{2}= -1

_{2 }= -1

_{2 }= ½

**36. The points B (1,3) and D (6,8) are two opposite vertices of a square ABCD. Find the equation of the diagonal AC.**

**Solution:**

_{1}= (8 – 3)/ (6 – 1) = 5/5 = 1

_{2}) will be

_{1 }× m

_{2 }= -1

_{2 }= -1

_{2}= -1

**37. ABCD is a rhombus. The co-ordinates of A and C are (3,6) and ( – 1,2) respectively. Write down the equation of BD.**

**Solution:**

_{1}) = (2 – 6)/ (-1 – 3) = -4/-4 = 1

_{2}

_{1}× m

_{2 }= -1

_{2 }= -1/(m

_{1})

**38. Find the equation of the line passing through the intersection of the lines 4x + 3y = 1 and 5x + 4y = 2 and**

**(i) parallel to the line x + 2y – 5 = 0**

**(ii) perpendicular to the x-axis.**

**Solution:**

**40. Find the image of the point (1,2) in the line x – 2y – 7 = 0.**

**Solution:**

_{1}) = ½

_{2}

_{1}× m

_{2 }= -1

_{2 }= -1

_{2 }= -2

**41. If the line x – 4y – 6 = 0 is the perpendicular bisector of the line segment PQ and the co-ordinates of P are (1,3), find the co-ordinates of Q.**

**Solution:**

**42. OABC is a square, O is the origin and the points A and B are (3,0) and (p,q). If OABC lies in the first quadrant, find the values of p and q. Also write down the equations of AB and BC.**

**Solution:**

^{2}+q

^{2}-6(3)=0⇒9+q

^{2}-18=0

^{2}-9=0⇒q

^{2}-9⇒q-3

**Chapter Test**

**1. Find the equation of a line whose inclination is 60° and y-intercept is – 4.**

**Solution:**

**2. Write down the gradient and the intercept on the y-axis of the line 3y + 2x = 12.**

**Solution:**

**3. If the equation of a line is y – √3x + 1, find its inclination.**

**Solution:**