What is the equation of general form that passes through the point 1 1 and has ay intercept of 2 xy 2 0x Y 2 0x Y 2 0?

A quadratic equation is a polynomial equation of degree 2 .  The standard form of a quadratic equation is

0 = a x 2 + b x + c

where   a , b and c are all real numbers and a ≠ 0 .

If we replace 0 with y , then we get a quadratic function

      y = a x 2 + b x + c

whose graph will be a parabola .

The axis of symmetry of this parabola will be the line x = − b 2 a . The axis of symmetry passes through the vertex, and therefore the x -coordinate of the vertex is − b 2 a . Substitute x = − b 2 a in the equation to find the y -coordinate of the vertex. Substitute few more x -values in the equation to get the corresponding y -values and plot the points. Join them and extend the parabola.

Example 1:

Graph the parabola y = x 2 − 7 x + 2 .

Compare the equation with y = a x 2 + b x + c to find the values of a , b , and c .

Here, a = 1 , b = − 7 and c = 2 .

Use the values of the coefficients to write the equation of axis of symmetry .

The graph of a quadratic equation in the form   y = a x 2 + b x + c has as its axis of symmetry the line x = − b 2 a . So, the equation of the axis of symmetry of the given parabola is x = − ( − 7 ) 2 ( 1 ) or x = 7 2 .

Substitute x = 7 2 in the equation to find the y -coordinate of the vertex.

y = ( 7 2 ) 2 − 7 ( 7 2 ) + 2         = 49 4 − 49 2 + 2         = 49   −   98   +   8 4           = − 41 4

Therefore, the coordinates of the vertex are ( 7 2 , − 41 4 ) .

Now, substitute a few more x -values in the equation to get the corresponding y -values.

Plot the points and join them to get the parabola.

Example 2:

Graph the parabola y = − 2 x 2 + 5 x − 1 .

Compare the equation with y = a x 2 + b x + c to find the values of a , b , and c .

Here, a = − 2 , b = 5 and c = − 1 .

Use the values of the coefficients to write the equation of axis of symmetry.

The graph of a quadratic equation in the form   y = a x 2 + b x + c has as its axis of symmetry the line x = − b 2 a . So, the equation of the axis of symmetry of the given parabola is x = − ( 5 ) 2 ( − 2 ) or x = 5 4 .

Substitute x = 5 4 in the equation to find the y -coordinate of the vertex.

y = − 2 ( 5 4 ) 2 + 5 ( 5 4 ) − 1         = − 50 16 + 25 4 − 1         = − 50   +   100   −   16 16           = 34 16           = 17 8

Therefore, the coordinates of the vertex are ( 5 4 , 17 8 ) .

Now, substitute a few more x -values in the equation to get the corresponding y -values.

Plot the points and join them to get the parabola.

Example 3:

Graph the parabola x = y 2 + 4 y + 2 .

Here, x is a function of y . The parabola opens "sideways" and the axis of symmetry of the parabola is horizontal. The standard form of equation of a horizontal parabola is x = a y 2 + b y + c where a , b , and c are all real numbers and   a ≠ 0 and the equation of the axis of symmetry is y = − b 2 a .

Compare the equation with x = a y 2 + b y + c to find the values of a , b , and c .

Here, a = 1 , b = 4 and c = 2 .

Use the values of the coefficients to write the equation of axis of symmetry.

The graph of a quadratic equation in the form   x = a y 2 + b y + c has as its axis of symmetry the line y = − b 2 a . So, the equation of the axis of symmetry of the given parabola is y = − 4 2 ( 1 ) or y = − 2 .

Substitute y = − 2 in the equation to find the x -coordinate of the vertex.

x = ( − 2 ) 2 + 4 ( − 2 ) + 2         = 4 − 8 + 2         = − 2

Therefore, the coordinates of the vertex are ( − 2 , − 2 ) .

Now, substitute a few more y -values in the equation to get the corresponding x -values.

Plot the points and join them to get the parabola.

We will learn how to find the equation of a straight line in intercept form.

The equation of a line which cuts off intercepts a and b respectively from the x and y axes is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1.

Let the straight line AB intersects the x-axis at A and the y-axis at B where OA = a and OB = b.

Now we have to find the equation of the straight line AB.

Let P(x, y) be any point on the line AB. Draw PQ perpendicular on OX and PR perpendicular on OX. Then, join the points O and P. Now, PQ = y, OQ = x.

Clearly, we see that

Area of the ∆OAB = Area of the ∆OPA + Area of the ∆OPB

⇒ ½ OA ∙ OB = ½ ∙ OA ∙ PQ + ½ ∙ OB ∙ PR

⇒ ½ a ∙ b = ½ ∙ a ∙ y + ½ ∙ b ∙ x

⇒ ab = ay + bx

⇒ \(\frac{ab}{ab}\) = \(\frac{ay + bx}{ab}\), dividing both sides by ab

⇒ 1 = \(\frac{ay}{ab}\) + \(\frac{bx}{ab}\)

⇒ 1 = \(\frac{y}{b}\) + \(\frac{x}{a}\)

⇒ \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1, which is the equation of the line in the intercept form.

The equation \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 is the satisfied by the co-ordinates of any point P lying on the line AB.

Therefore, \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 represent the equation of the straight line AB.

Solved examples to find the equation of a straight line in intercept form:

1. Find the equation of the line which cuts off an intercept 3 on the positive direction of x-axis and an intercept 5 on the negative direction of y-axis.

Solution: 

The equation of a line which cuts off intercepts a and b respectively from the x and y axes is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1.

Here, a = 3 and b = -5

Therefore, the equation of the straight line is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 ⇒ \(\frac{x}{3}\) + \(\frac{y}{-5}\) = 1 ⇒ \(\frac{x}{3}\) - \(\frac{y}{5}\) = 1 ⇒ 5x – 3y = 15 ⇒ 5x – 3y – 15 = 0.

2. Find the intercepts of the straight line 4x + 3y = 24 on the co-ordinate axes.

Solution:

Given equation 4x + 3y = 24.

Now convert the given equation into intercept form.

4x + 3y = 24

⇒ \(\frac{4x + 3y}{24}\) = \(\frac{24}{24}\), Dividing both sides by 24

⇒ \(\frac{4x}{24}\) + \(\frac{3y}{24}\) = 1

⇒ \(\frac{x}{6}\) + \(\frac{y}{8}\) = 1, which is the intercept form.

Therefore, x-intercept = 6 and y-intercept = 8.

Note: (i) The straight line \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 intersects the x-axis at A(a, 0) and the y-axis at B(0, b).

(ii) In \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1, a is x-intercept and b is y- intercept.

These intercept a and b may be positive as well as negative.

(iii) If the straight line AB passing through the origin then, a = 0 and b = 0. If we put a = 0 and b = 0 in the intercept form, then \(\frac{x}{0}\) + \(\frac{y}{0}\) = 1, which is undefined. For this reason the equation of a straight line passing through the origin cannot be expressed in the intercept form.

(iv) A line parallel to the x-axis does not intercept the x-axis at any finite distance and hence, we cannot get any finite x- intercept (i.e., a) of such a line. For this reason, a line parallel to x-axis cannot be expressed in the intercept from. In like manner, we cannot get any finite y- intercept (i.e., b) of a line parallel to y-axis and hence, such a line cannot be expressed in the intercept form.

● The Straight Line

11 and 12 Grade Math

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