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The Cartesian equation of the line passing through the point (1, 2, -1) and parallel to the line 5x−25=14−7y=35z5x - 25 = 14 - 7y = 35z is

Solution

✅ Correct Option: 2

The line must pass through point (1, 2, -1) and be parallel to the line 5x−25=14−7y=35z5x - 25 = 14 - 7y = 35z.

Parallel lines have the same direction ratios.


The given line is: 5x−25=14−7y=35z5x - 25 = 14 - 7y = 35z

Setting each expression equal to parameter kk:

5x−25=14−7y=35z=k5x - 25 = 14 - 7y = 35z = k


From 5x−25=k5x - 25 = k:

5x=k+255x = k + 25

x=k5+5x = \frac{k}{5} + 5

From 14−7y=k14 - 7y = k:

−7y=k−14-7y = k - 14

y=−k7+2y = -\frac{k}{7} + 2

From 35z=k35z = k:

z=k35z = \frac{k}{35}


The coefficients of kk give direction ratios:

15:−17:135\frac{1}{5} : -\frac{1}{7} : \frac{1}{35}

Multiplying by 35 (the LCM):

35×15:35×(−17):35×13535 \times \frac{1}{5} : 35 \times \left(-\frac{1}{7}\right) : 35 \times \frac{1}{35}

=7:−5:1= 7 : -5 : 1

Direction ratios are (7, -5, 1)


The line passes through (1, 2, -1) with direction ratios (7, -5, 1).

Using the standard form:

x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}

where (x1,y1,z1)=(1,2,−1)(x_1, y_1, z_1) = (1, 2, -1) and (a,b,c)=(7,−5,1)(a, b, c) = (7, -5, 1)

x−17=y−2−5=z−(−1)1\frac{x - 1}{7} = \frac{y - 2}{-5} = \frac{z - (-1)}{1}

x−17=y−2−5=z+11\frac{x - 1}{7} = \frac{y - 2}{-5} = \frac{z + 1}{1}

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