11th Grade Geometry

Introduction of 11th grade geometry :-

In grade (11) means eleventh grade is a year of education in all over the world. The eleventh grade is the final year of the secondary school. Students are usually 16 - 17 years old. Geometry is concerned with size, shape, relative figures etc. In 11th grade geometry lessons study about the angle, circle , quadrilateral etc.

Example Problems for 11th Grade Geometry :-

Problem 1:-

Determine the equation of the straight line passing through the points (1, 2) and (3, − 4).

Solution:

The equation of a straight line passing through two points is

`(y - y1)/ (y1 - y2)` = `(x - x1)/( x1 - x2)`


Here (x1, y1) = (1, 2) and (x2, y2) = (3, − 4).

Substituting the above, the required line is

`(y - 2)/(2 + 4)` = `(x - 1)/(1 - 3)`

`(y - 2)/6` = `(x - 1)/(- 2)`

`(y-2)/3` = `(x-1)/(-1)`

y − 2 = − 3 (x − 1)
y − 2 = − 3x + 3

3x + y = 5 is the required equation of the straight line.Is this topic AAA Postulate hard for you? Watch out for my coming posts.

Problem 2:-

Find the equation of the straight line passing through the point (1,2) and making intercepts on the co-ordinate axes which are in the ratio 2 : 3.

Solution:-

The intercept form is

`x/a +y/b` = 1 … (1)

The intercepts are in the ratio 2 : 3  a = 2k, b = 3k.

(1) becomes

`x/(2k) +y/(3k)` = 1     i.e. 3x + 2y = 6k

Since (1, 2) lies on the above straight line, 3 + 4 = 6k i.e. 6k = 7

Hence the required equation of the straight line is 3x + 2y = 7


Problem 3:-

Find the distance between the parallel lines 2x + 3y − 6=0 and 2x + 3y + 7 = 0.

Solution:-

The distance between the parallel lines is

`|(c_1 - c_2)/sqrt(a^2 + b^2)|` .

Here `c_1` = − 6, `c_2` = 7, a = 2, b = 3

The required distance is

`|(- 6 -7)/sqrt(2^2 + 3^2)|` = `| (-13)/sqrt(13)|`

= `sqrt(13)` units.


Practice Problems for 11th Grade Geometry :-

Problem 1:-

Find the equation of the straight line, if the perpendicular from the origin makes an angle of 120° with x-axis and the length of theperpendicular from the origin is 6 units.

Answer: The required equation of the straight line is x − `sqrt(3)` y + 12 = 0


Problem 2:-

Find the points on y-axis whose perpendicular distance from the straight line 4x − 3y − 12 = 0 is 3.

Answer: The required points are (0, 1) and (0, − 9).