7.) Write the equation 3x + 4y – 7 = 0 in normal form.
3/5x + 4/5y + 7/5 = 0
-3/5x - 4/5y - 7/5 = 0
-3/5x - 4/5y + 7/5 = 0 ~
3/5x + 4/5y - 7/5 = 0
8.)Write the standard form of the equation of a line for which the length of the normal is 4 and the normal makes an angle of 45° with the positive x–axis.
sqrt2x + sqrt2y +8 = 0
2x + 2y – 8 = 0
sqrt2x + sqrt2y -8 = 0 ~
2x + 2y + 8 = 0
9.) Write the standard form of the equation of a line for which the length of the normal is 3 and the normal makes an angle of 135° with the positive x–axis.
sqrt2^x – sqrt2y + 6 = 0 ~
sqrt2^x + sqrt2y - 6 = 0
sqrt2^x – sqrt2y - 6 = 0
sqrt2^x + sqrt2y + 6 = 0
10.) Find the distance between P (–2, 1) and the line with equation x – 2y + 4 = 0.
4sqrt5/5
0 ~
-4sqrt5/5
5/4
I agree with all your answers
huihjn
To find the distance between a point and a line, we can use the formula:
Distance = |Ax + By + C| / sqrt(A^2 + B^2)
Given the line equation x - 2y + 4 = 0, we can rewrite it in the general form Ax + By + C = 0:
1x - 2y + 4 = 0
Comparing this with Ax + By + C = 0, we have A = 1, B = -2, and C = 4.
Now, let's find the distance between the point P(-2, 1) and the line:
Distance = |1(-2) + (-2)(1) + 4| / sqrt(1^2 + (-2)^2)
= |-2 - 2 + 4| / sqrt(1 + 4)
= |0| / sqrt(5)
= 0 / sqrt(5)
= 0
So, the distance between P(-2, 1) and the line x - 2y + 4 = 0 is 0.
To write the equation 3x + 4y - 7 = 0 in normal form, you need to rearrange the terms so that the coefficients of x and y are fractions with a common denominator. The equation can be rewritten as:
3/5x + 4/5y - 7/5 = 0
For the equation of a line in standard form where the length of the normal is 4 and the normal makes an angle of 45° with the positive x-axis, you can use the formula:
ax + by + c = 0
The length of the normal is the coefficient of x or y divided by the square root of the sum of the squares of the coefficients, so:
4 = sqrt(a^2 + b^2)
Since the angle with the positive x-axis is 45°, the slope of the line is 1, which means a = b. Substituting this into the formula, you get:
4 = sqrt(2a^2)
Solving for a, you get:
a = 2
Therefore, the standard form of the equation is:
2x + 2y - 8 = 0
Similarly, for the equation of a line where the length of the normal is 3 and the normal makes an angle of 135° with the positive x-axis, you can apply the same process. The coefficient of x is √2, and since the angle is 135°, the slope of the line is -1, so the coefficient of y is -√2. This gives you the equation:
√2x - √2y + 6 = 0
To find the distance between point P(-2, 1) and the line x - 2y + 4 = 0, you can use the formula for the perpendicular distance from a point to a line. The formula is:
Distance = |ax0 + by0 + c| / sqrt(a^2 + b^2)
Substituting the values, you get:
Distance = |(-2) - 2(1) + 4| / sqrt(1^2 + (-2)^2)
= |-2 - 2 + 4| / sqrt(1 + 4)
= |-2| / sqrt(5)
= 2 / sqrt(5)
= 2sqrt(5) / 5
Therefore, the distance between point P and the line is 2sqrt(5) / 5.