Maths Question 3

Distance Section Formula

Q.1. The mid point of the line segment joining (2a, 4) and (– 2, 2b) is
(1, 2a + 1). Find the value of a and b.

Solution :

Mid point of (2a, 4) and (– 2, 2b) is (1, 2a + 1)
i.e. x = (x₁ + x₂)/2 Or, 1 = (2a – 2)/2 => a = 2;
and y = (y₁ + y₂)/2 Or, 2a + 1 = (4 + 2b)/2 => 2a + 1 = 2 + b => b = 3;
Hence, a = 2, b = 3. [Ans.]

Q.2. If the points (2,1) and (1, – 2) are equidistant from the point (x,y), show that x + 3y = 0.

Solution :

Let A(2,1) and B(1, – 2) is equidistant from P(x,y), then PA = PB
Or, √[(x – 2)² + (y – 1)²] = √[(x – 1)² + (y +2)²] [by distance formula]
Or, x² – 4x + 4 + y² – 2y + 1 = x² – 2x + 1 + y² + 4y + 4
Or, – 4x – 2y + 2x – 4y = 0
Or, – 2x – 6y = 0 Or, x + 3y = 0 [Proved.]

Q.3. Find the ratio in which the point (1 ,– 1) divides the line joining the points (a, 1) and (11, 4). Also find the value of a.

Solution :

Let line joining the points (a,1) and (11,4) is divided by the point (1, – 1) in the ratio K : 1, then
1 = (1×a + K×11)/(K+1) ------------------ (i)
and –1 = (1×1+K×4)/(K+1) ------------------ (ii)
From (ii) we get 4K + 1 = – K – 1
Or, 5K = – 2 Or, K = – 2/5
Putting the value of K in (i) we get,
11K + a = K + 1 or, 10K + a = 1
or, 10 (– 2 /5) + a = 1 or, a = 1 + 4 = 5. [Ans.]

Q.4. Find the value of ‘k’ if the triangle formed by A(8, – 10), B(7, – 3) and C(0,k) is right angled at B.

Solution :

As B is a right angle, so, AC must be the hypotenuse and hence by
Pythagoras Theorem, AC² = AB² + BC²
Or, (0 – 8)² + (k + 10)² = (7 – 8)²+ (– 3 + 10)² + (0 – 7)² + (k +3)²
Or, 64 + k² + 20k + 100 = 1 + 49 + 49 + k² + 6k + 9
Or, 14k = 108 – 164 = – 56 Or, k = – 56 /14 = – 4. [Ans.]

Equation of a Straight Line

Q.1. Find the equation of a line passing through (2, –3) and inclined at an angle of 135º with positive direction of x-axis.

Solution :

The point is (x1,y1) = (2, –3), slope = m = tan135º = –1,
The equation of a line passing through (x1,y1) with slope m is :
y – y1 = m(x – x1)
Hence, equation of line passing through (2, –3) with slope –1 is given by
y – (–3) = –1(x - 2)
Or, y + 3 = –x + 2
Or, x + y + 1 = 0. [Ans.]

Q.2. Find the equation of the line parallel to 3x + 2y = 8 and passing through the point (0, 1).

Solution :

The equation of line parallel to 3x + 2y = 8 may be written as,
3x + 2y = k -------- (i) we have to find the value of k.
Line (i) passes through (0, 1), therefore, 3(0) + 2(1) = k
Or, k = 2,
Hence, equation of required line is 3x + 2y = 2. [Ans.]

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

Solution :

Do yourself. [Ans. 3x + 5y – 20 = 0.]

Q.4. The line 4x – 3y + 12 = 0 meets the x-axis at A. Write down the co-ordinates of A. Determine the equation of the line passing through A and perpendicular to 4x – 3y + 12 = 0.

Solution :

At x-axis, y-co-ordinate is zero. The line 4x – 3y + 12 = 0, meets x-axis. Hence putting y = 0 in the equation of line we get,
4x – 3×0 + 12 = 0 Or, 4x = –12 or, x = –3.
Therefore, line 4x – 3y + 12 = 0, meets x-axis at A(–3,0).[Ans.]
Slope m₁ of the line 4x – 3y + 12 = 0, = – (coefficient of x /coefficient of y)
= – (4/–3) = 4/3
Let the slope of the required line be m2.
Then m₁×m₂ = –1
Or, (4/3) ×m₂ = –1 => m₂ = –3/4
Therefore, line through A(–3,0 ) with slope –3/4 is given by
y – 0 = (–3/4){x – (–3)}
Or, 4y = –3(x + 3) = –3x – 9
Or, 3x + 4y + 9 = 0. [Ans.]

Linear Inequation in One Unknown

Q.1. Solve the following inequation and graph the solution on the number line.
– 8/3 ≤ x + 1/3 <>

Solution :

The given inequation has two parts :
– 8/3 ≤ x + 1/3 and x + 1/3 < 10/3
Or, – 8/3 – 1/3 ≤ x and x < 10/3 – 1/3
Or, – 9/3 ≤ x and x < 9/3
Or, – 3 ≤ x and x < 3
Or, – 3 ≤ <>Ans.]
The required graph is :

Fig.

Q.2. Find the range of values of x, which satisfy the inequation
– 1/5 ≤ 3x/10 + 1 <>Graph the solution set on the number line.

Solution :

Here we have, - 1/5 ≤ 3x/10 + 1 < 2/5 .
Multiplying throughout by LCM of 5, 10, 5 i.e. 10 we get ,
– 2 ≤ 3x + 10 < 4
Adding – 10 to both sides , we get

1. 2 – 10 ≤ 3x + 10 + (– 10) <>
2. 12 ≤ 3x < – 6
3. 4 ≤ x < – 2 [dividing throughout by 3]

Hence, the solution set is {x: x ε R, – 4 ≤ x < – 2 }.[Ans.]


Here, – 4 is included and – 2 is not included.

Q.3. Solve the following inequation, and graph the solution on the number line :
2x – 5 ≤ 5x + 4 <>

Solution :

Here we have, 2x – 5 ≤ 5x + 4 < 11
i.e. 2x – 5 ≤ 5x + 4 and 5x + 4 < 11
or, 2x – 5 + (– 5x + 5) ≤ 5x + (– 5x + 5) and 5x + 4 + (– 4 ) < 11 + (– 4 )
or, 0 – 3x ≤ 9 and 5x < 7
or, x ≥ – 3 and x < 7/5 or, 0 – 3 ≤ x and x < 7/5
Hence, – 3 ≤ x < 7/5, x ε R.
Therefore, solution set is {x : x ε R, – 3 ≤ x <>Ans.]