11类NCERT解决方案-第10章直线–练习10.2 |套装2

第10章直线–练习10.2 |套装1

问题11.垂直于连接点(1、0)和(2、3)的线段的线将其按比例1：n划分。找到直线的方程式。

解决方案：

Slope of line segment will be m1 = (y2 – y1)/(x2 – x1)

= (3 – 0)/(2 – 1)

= 3/1

= 3

If two line are perpendicular to each other than slope of line1 X slope of line2 = -1

So, the slope of line will be m2 = (-1/m1)

m2 = -1/3

As we know that the coordinates of a point (p, q) dividing the line 

segment joining the points (x1, y1) and (x2, y2) internally 

In the ratio m : n are:

(p, q) = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))

(p, q) = ((1 × 2 + n × 1)/(1 + n), (1 × 3 + n × 0)/(1 + n))

p = (2 + n)/(1 + n)

q = 3/(1 + n)

We know that the point (p, q) lies on the line with slope m2,

Equation of line will be y – q = m2(x – p)

y – 3/(1 + n) = (–1/3)(x – (2 + n)/(1 + n))

3((1 + n) y – 3) = (–(1 + n) x + 2 + n)

3(1 + n) y – 9 = – (1 + n) x + 2 + n

(1 + n) x + 3(1 + n) y – n – 9 – 2 = 0

(1 + n) x + 3(1 + n) y – n – 11 = 0

So, the equation of the line is (1 + n) x + 3(1 + n) y – n – 11 = 0

问题12。找到一条线的方程，该方程在坐标轴上切掉相等的截距并通过点(2，3)。

解决方案：

Given that the line cuts off equal intercepts on the coordinate axes i.e. a = b.

So, the equation of the line intercepts a and b on x-and y-axis, respectively, which is

x/a + y/b = 1

So, x/a + y/a = 1

x + y = a            -(1)

Given point is (2, 3)

2 + 3 = a

a = 5

On substituting the value of ‘a’ in eq(1), we get

x + y = 5

x + y – 5 = 0

So, the equation of the line is x + y – 5 = 0.

问题13。求出通过点(2，2)的线的方程，并截去总和为9的轴上的截距。

解决方案：

Equation of the line making intercepts a and b on x-and y-axis, respectively, is 

x/a + y/b = 1          -(1)

Given that sum of intercepts = 9

a + b = 9

b = 9 – a

On substituting the value of b in the eq(1), we get

x/a + y/(9 – a) = 1

Given that the line passes through the point (2, 2),

So, 2/a + 2/(9 – a) = 1

[2(9 – a) + 2a] / a(9 – a) = 1

[18 – 2a + 2a] / a(9 – a) = 1

18/a(9 – a) = 1

18 = a (9 – a)

18 = 9a – a²

a² – 9a + 18 = 0

By using factorization method for quadratic equation solving, we get

a2 – 3a – 6a + 18 = 0

a (a – 3) – 6 (a – 3) = 0

(a – 3) (a – 6) = 0

a = 3 or a = 6

Let us substitute in eq(1),

Case 1 (a = 3):

Then b = 9 – 3 = 6

x/3 + y/6 = 1

2x + y = 6

2x + y – 6 = 0

Case 2 (a = 6):

Then b = 9 – 6 = 3

x/6 + y/3 = 1

x + 2y = 6

x + 2y – 6 = 0

So, the equation of the line is 2x + y – 6 = 0 or x + 2y – 6 = 0

问题14.找到通过点(0，2)与正x轴成2π/ 3角的直线方程。另外，找到与之平行且与y轴相交的直线方程，该方程在原点以下2个单位的距离处。

解决方案：

Given that

Point (0, 2) and θ = 2π/3

slope (m)= tan θ

m = tan (2π/3) = -√3

Now, the equation of line passing through point(p,q) with slope m will be:

y – q = m (x – p)

y – 2 = -√3 (x – 0)

y – 2 = -√3 x

√3 x + y – 2 = 0

Now we need to find equation of line parallel to above obtained equation 

crosses the y-axis at a distance of 2 units below the origin.

So, the point = (0, -2) and for parallel line slope will be same that is m = -√3

From point slope form equation,

y – (–2) = –√3 (x – 0)

y + 2 = -√3 x

√3 x + y + 2 = 0

So, the equation of line is √3 x + y – 2 = 0 and the line parallel to it is √3 x + y + 2 = 0.

问题15.从原点到直线的垂直线在点(–2，9)处与直线相交，找到直线的方程。

解决方案：

Perpendicular line will pass through point (0,0) and (-2,9) as per given question,

Then slope of perpendicular line will be m1 = (y2 – y1)/(x2 – x1)

= (9 – 0)/(–2 – 0)

= –9/2

As we know that two non-vertical lines are perpendicular to each 

other if their slopes are negative reciprocals of each other.

So, we need to find equation of line let say it AB.

Slope of line AB will be m = (–1/m1) = –1/(–9/2) = 2/9

As perpendicular line and line intersecting at (–2, 9) that means (–2, 9) lies on the line AB

By using Slope Point Form of line, we get

y – y1 = m (x – x1)

y – 9 = (2/9) (x – (–2))

9(y – 9) = 2(x + 2)

9y – 81 = 2x + 4

2x + 4 – 9y + 81 = 0

2x – 9y + 85 = 0

So, the equation of line is 2x – 9y + 85 = 0.

问题16.铜棒的长度L(厘米)是其摄氏温度C的线性函数。在实验中，如果当C = 20时L = 124.942，而当C = 110时L = 125.134，则用C。

解决方案：

Let us considered ‘L’ along X-axis and ‘C’ along Y-axis, 

So, we have two points (124.942, 20) and (125.134, 110) in XY-plane.

As we know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by

$y - y1 = \frac{(y2 - y1)}{(x2 - x1)}(x -x1)$

$C-20 = \frac{(110-20)}{(125.134-124.942)}(L-124.942)$

0.192(C-20) = 90(L – 124.942)

 L can be express in Terms of C in following manner

L = (0.192 × (C – 20)/90) + 124.942

问题17.一家牛奶店的老板发现他每周可以卖980卢比的牛奶。每周每公升14升牛奶和1220升牛奶为卢比。 16 /升假设售价与需求之间存在线性关系，则他每周可以以卢比出售多少公升汽油。 17 /升?

解决方案：

Let us considered the relationship between selling price and demand is linear.

So, the selling price per liter along X-axis and demand along Y-axis, 

We have two points (14, 980) and (16, 1220) in XY-plane

We know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by

$y - y1 = \frac{(y2 - y1)}{(x2 - x1)}(x -x1)$

$y-980 =\frac{(1220-980)}{2}(x-14)$

y – 980 = 120 (x – 14)

y = 120 (x – 14) + 980

When x = Rs 17/liter,

y = 120 (17 – 14) + 980

y = 120(3) + 980

y = 360 + 980 = 1340

So, the owner can sell 1340 liters weekly at Rs. 17/liter.

问题18. P(a，b)是轴之间的线段的中点。显示直线方程为x / a + y / b = 2

解决方案：

Let us considered AB be a line segment whose midpoint is P (a, b).

So, the coordinates of A is (0, y) and B is (x, 0) 

Mid-Point of AB will be P(a, b) = ((x + 0)/2, (y + 0)/2)

a = x/2 

x = 2a

And

b = y/2 

y = 2b

That means line AB is passing from (0, 2b) and (2a, 0)

We know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by

$y - y1 = \frac{(y2 - y1)}{(x2 - x1)}(x -x1)$

(y – 2b) = ((0 – 2b)/2a) × x

a (y – 2b) = –bx

ay – 2ab = –bx

bx + ay = 2ab

On dividing both the sides with ab, we get

x/a + y/b =2

问题19.点R(h，k)将轴之间的线段按比例1：2划分。找到线的方程。

解决方案：

Let us consider, PQ be the line segment such that r (h, k) divides it in the ratio 1: 2.

So the coordinates of P and Q be (0, y) and (x, 0) respectively.

We know that the coordinates of a point dividing the line segment

 joins the points (x1, y1) and (x2, y2) internally in the ratio m: n is

(h, k) = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))

(h, k) = ((1 × 0 + 2 × x)/(1 + 2), (1 × y + 2 × 0)/(1 + 2))

h = 2x/3 and k = y/3

x = 3h/2 and y = 3k

So, P = (0, 3k) and Q = (3h/2, 0)

We know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by

$y - y1 = \frac{(y2 - y1)}{(x2 - x1)}(x -x1)$

y – 3k = ((0 – 3k)/(3h/2))x

3h(y – 3k) = –6kx

3hy – 9hk = –6kx

6kx + 3hy = 9hk

On dividing both the sides by 9hk, we get,

2x/3h + y/3k = 1

So, the equation of the line is given by 2x/3h + y/3k = 1

问题20.通过使用直线方程的概念，证明三个点(3，0)，(– 2，– 2)和(8，2)是共线的。

解决方案：

Prove that the given three points (3, 0), (– 2, – 2) and (8, 2) are collinear, 

So, we have to also prove that the line passing through the points 

(3, 0) and (– 2, – 2) also passes through the point (8, 2).

The equation of the line passing through the points (x1, y1) and (x2, y2) is given by

$y - y1 = \frac{(y2 - y1)}{(x2 - x1)}(x -x1)$

y – 0 = ((–2 – 0)/(–2 – 3)) × (x – 3)

–5y = –2 (x – 3)

–5y = –2x + 6

2x – 5y = 6

Checking whether (8,2) is on line 2x – 5y = 6 or not,

LHS = 2x – 5y = 2(8) – 5(2)

= 16 – 10

= 6

= RHS

So, the line passing through the points (3, 0) and (– 2, – 2) 

also passes through the point (8, 2).

Hence, proved that the points (3, 0), (– 2, – 2) and (8, 2) are collinear,